Question 1: Palette 1/107

DPT MCQ

Inverse trigonometric identity: $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$ for $x \in [-1, 1]$.
Basic integration rule: $\int k \, dx = kx + C$, where $k$ is a constant.

Solve $\int (\sin^{-1}x + \cos^{-1}x) \, dx$

A

$\frac{1}{2}\pi x + c$

B

$x(\sin^{-1}x + \cos^{-1}x) + c$

C

$x(\cos^{-1}x + \sin^{-1}x) + c$

D

$\frac{\pi}{2} + x + c$

Question 2: Palette 2/107

DPT MCQ

Logarithmic Identity: $e^{a\log x} = e^{\log(x^a)} = x^a$
Power Rule of Integration: $\int x^n \, dx = \frac{x^{n+1}}{n+1} + c$ (for $n \neq -1$)

$\int \frac{e^{5\log x} - e^{4\log x}}{e^{3\log x} - e^{2\log x}} \, dx =$

A

$e \cdot 3^{-3x} + c$

B

$e^{3\log x} + c$

C

$\frac{x^3}{3} + c$

D

None of these

Question 3: Palette 3/107

DPT MCQ

Rationalization of Irrational Denominators: $\frac{1}{\sqrt{a} + \sqrt{b}} = \frac{\sqrt{a} - \sqrt{b}}{(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b})} = \frac{\sqrt{a} - \sqrt{b}}{a - b}$
Power Rule of Integration: $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$

$\int \frac{dx}{\sqrt{1 + x} + \sqrt{x}} =$

A

$\frac{2}{3}(1+x)^{\frac{3}{2}} - \frac{2}{3}x^{\frac{3}{2}} + c$

B

$\frac{3}{2}(1+x)^{\frac{2}{3}} + \frac{3}{2}x^{\frac{2}{3}} + c$

C

$\frac{3}{2}(1+x)^{\frac{3}{2}} + \frac{3}{2}x^{\frac{3}{2}} + c$

D

$\frac{2}{3}(1+x)^{\frac{3}{2}} - \frac{2}{3}x^{\frac{3}{2}} + c$

Question 4: Palette 4/107

DPT MCQ

Algebraic Factorization Identity: $x^4 + x^2 + 1 = (x^2 + x + 1)(x^2 - x + 1)$
Power Rule of Integration: $\int x^n \, dx = \frac{x^{n+1}}{n+1} + c$

$\int \frac{x^4 + x^2 + 1}{x^2 - x + 1} \, dx =$

A

$\frac{1}{3}x^3 + \frac{1}{2}x^2 + x + c$

B

$\frac{1}{3}x^3 - \frac{1}{2}x^2 + x + c$

C

$\frac{1}{3}x^3 + \frac{1}{2}x^2 - x + c$

D

None of these

Question 5: Palette 5/107

DPT MCQ

Double-angle Trigonometric Identity: $\cos 2\theta = 2\cos^2 \theta - 1$
Difference of Squares: $a^2 - b^2 = (a - b)(a + b)$
Basic Integration Rules: $\int \cos x \, dx = \sin x + c$ and $\int k \, dx = kx + c$

$\int \frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha} \, dx =$

A

$2[\sin x + x \cos \alpha] + c$

B

$2[\sin x + \sin \alpha] + c$

C

$2[-\sin x + x \cos \alpha] + c$

D

$-2[\sin x + \sin \alpha] + c$

Question 6: Palette 6/107

DPT MCQ

Algebraic Identity: $(a - b)^2 = a^2 + b^2 - 2ab$
Trigonometric Identities: $\tan x \cdot \cot x = 1$, $\tan^2 x = \sec^2 x - 1$, and $\cot^2 x = \csc^2 x - 1$
Standard Integrals: $\int \sec^2 x \, dx = \tan x + c$ and $\int \csc^2 x \, dx = -\cot x + c$

$\int (\tan x - \cot x)^2 \, dx =$

A

$\tan x + \cot x + c$

B

$\sec x \tan x + c$

C

$\csc x \cot x + c$

D

None of these

Question 7: Palette 7/107

DPT MCQ

Algebraic Identity: $(a + b)^2 = a^2 + b^2 + 2ab$
Trigonometric Identity: $\tan^2 x = \sec^2 x - 1$
Standard Integrals: $\int \sec^2 x \, dx = \tan x + c$ and $\int \sec x \tan x \, dx = \sec x + c$

$\int (\sec x + \tan x)^2 \, dx =$

A

$2(\sec x + \tan x) - x + c$

B

$\frac{1}{3}(\sec x + \tan x)^3 + c$

C

$\sec x(\sec x + \tan x) + c$

D

$2(\sec x + \tan x) + c$

Question 8: Palette 8/107

DPT MCQ

Rationalization of Trigonometric Denominators: $\frac{\csc\theta - \cot\theta}{\csc\theta + \cot\theta} = \frac{(\csc\theta - \cot\theta)^2}{\csc^2\theta - \cot^2\theta}$
Trigonometric Identity: $\csc^2\theta - \cot^2\theta = 1$
Expansion and Identity: $(\csc\theta - \cot\theta)^2 = \csc^2\theta + \cot^2\theta - 2\csc\theta\cot\theta = 2\csc^2\theta - 2\csc\theta\cot\theta - 1$
Standard Integrals: $\int \csc^2\theta \, d\theta = -\cot\theta + c$ and $\int \csc\theta\cot\theta \, d\theta = -\csc\theta + c$

$\int \frac{\csc\theta - \cot\theta}{\csc\theta + \cot\theta} \, d\theta =$

A

$2\csc\theta - 2\cot\theta - \theta + c$

B

$2\csc\theta - 2\cot\theta + \theta + c$

C

$2\csc\theta + 2\cot\theta - \theta + c$

D

None of these

Question 9: Palette 9/107

DPT MCQ

Rationalization of Trigonometric Denominators: $\frac{1}{1 - \sin x} = \frac{1 + \sin x}{(1 - \sin x)(1 + \sin x)} = \frac{1 + \sin x}{\cos^2 x}$
Trigonometric Identities: $\frac{1}{\cos^2 x} = \sec^2 x$ and $\frac{\sin x}{\cos^2 x} = \tan x \sec x$
Standard Integrals: $\int \sec^2 x \, dx = \tan x + c$ and $\int \sec x \tan x \, dx = \sec x + c$

$\int \frac{dx}{1 - \sin x} =$

A

$x + \cos x + c$

B

$1 + \sin x + c$

C

$\sec x - \tan x + c$

D

$\sec x + \tan x + c$

Question 10: Palette 10/107

DPT MCQ

Half-Angle Trigonometric Identity: $1 + \cos x = 2\cos^2\left(\frac{x}{2}\right)$
Standard Integration Formula: $\int \cos(kx) \, dx = \frac{\sin(kx)}{k} + c$

$\int \sqrt{1 + \cos x} \, dx \text{ equals}$

A

$2\sqrt{2}\sin\frac{x}{2} + c$

B

$-2\sqrt{2}\sin\frac{x}{2} + c$

C

$-2\sqrt{2}\cos\frac{x}{2} + c$

D

$2\sqrt{2}\cos\frac{x}{2} + c$

Question 11: Palette 11/107

DPT MCQ

Trigonometric Half-Angle Identities: $1 = \cos^2\left(\frac{x}{2}\right) + \sin^2\left(\frac{x}{2}\right)$ and $\sin x = 2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)$
Square of a Binomial Identity: $\cos^2\left(\frac{x}{2}\right) + \sin^2\left(\frac{x}{2}\right) + 2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right) = \left(\cos\left(\frac{x}{2}\right) + \sin\left(\frac{x}{2}\right)\right)^2$
Standard Integration Formula: $\int \sin(kx) \, dx = -\frac{\cos(kx)}{k} + c$ and $\int \cos(kx) \, dx = \frac{\sin(kx)}{k} + c$

$\int \sqrt{1 + \sin x} \, dx =$

A

$\frac{1}{2}\left(\sin\frac{x}{2} + \cos\frac{x}{2}\right) + c$

B

$\frac{1}{2}\left(\sin\frac{x}{2} - \cos\frac{x}{2}\right) + c$

C

$2\sqrt{1 + \sin x} + c$

D

$-2\sqrt{1 - \sin x} + c$

Question 12: Palette 12/107

DPT MCQ

Trigonometric Half-Angle Identity: $\cos^2 x = \frac{1 + \cos 2x}{2}$
Standard Integrals: $\int 1 \, dx = x + c$ and $\int \cos(kx) \, dx = \frac{\sin(kx)}{k} + c$

$\int \cos^2 x \, dx =$

A

$\frac{1}{2}\left(x + \frac{\sin 2x}{2}\right) + c$

B

$\frac{1}{2}\left(x - \frac{\sin 2x}{2}\right) + c$

C

$\frac{\sin^3 x}{3} + c$

D

$-\frac{1}{2}\sin 2x + c$

Question 13: Palette 13/107

DPT MCQ

Triple-Angle Trigonometric Identity: $\cos 3x = 4\cos^3 x - 3\cos x \implies \cos^3 x = \frac{\cos 3x + 3\cos x}{4}$
Standard Integration Formula: $\int \cos(kx) \, dx = \frac{\sin(kx)}{k} + c$

$\int \cos^3 x \, dx =$

A

$\frac{1}{4}\left(\sin 3x + 3\sin x\right) + c$

B

$\frac{1}{4}\left(\frac{\sin 3x}{3} + 3\sin x\right) + c$

C

$\frac{\cos^4 x}{4} + c$

D

$\frac{1}{3}\left(\sin 3x + \cos x\right) + c$

Question 14: Palette 14/107

DPT MCQ

Power Reduction Identity: $\cos^2 x = \frac{1 + \cos 2x}{2}$
Square of a Binomial: $(1 + \cos 2x)^2 = 1 + 2\cos 2x + \cos^2 2x$
Double-Angle Substitution: $\cos^2 2x = \frac{1 + \cos 4x}{2}$

$\int \cos^4 x \, dx =$

A

$\frac{3x}{8} + \frac{\sin 2x}{4} + \frac{\sin 4x}{32} + c$

B

$\frac{x}{4} + \frac{\sin 2x}{2} + \frac{\sin 4x}{8} + c$

C

$\frac{\sin^5 x}{5} + c$

D

None of these

Question 15: Palette 15/107

DPT MCQ

Product-to-Sum Trigonometric Identity: $2\cos A \cos B = \cos(A + B) + \cos(A - B)$
Standard Integration Formula: $\int \cos(kx) \, dx = \frac{\sin(kx)}{k} + c$

$\int \cos 2x \cos 3x \, dx =$

A

$\frac{1}{2}\left( \frac{\sin 5x}{5} + \sin x \right) + c$

B

$\frac{1}{2}\left( \frac{\sin 5x}{5} - \sin x \right) + c$

C

$\frac{1}{2}\left( \cos 5x + \cos x \right) + c$

D

$\frac{\sin 2x \sin 3x}{6} + c$

Question 16: Palette 16/107

DPT MCQ

Double-Angle Identity: $\sin x \cos x = \frac{\sin 2x}{2} \implies \sin^3 x \cos^3 x = \frac{\sin^3 2x}{8}$
Triple-Angle Identity: $\sin 3\theta = 3\sin\theta - 4\sin^3\theta \implies \sin^3\theta = \frac{3\sin\theta - \sin 3\theta}{4}$

$\int \sin^3 x \cos^3 x \, dx =$

A

$\frac{1}{32}\left( -\frac{3\cos 2x}{2} + \frac{\cos 6x}{6} \right) + c$

B

$\frac{1}{8}\left( \frac{\sin^4 x}{4} \right) + c$

C

$\frac{1}{32}\left( \frac{3\cos 2x}{2} - \frac{\cos 6x}{6} \right) + c$

D

None of these

Question 17: Palette 17/107

DPT MCQ

Trigonometric Identity: $\cos^2 x = 1 - \sin^2 x$
Algebraic Splitting: $\frac{x^2 + 1 - \sin^2 x}{1 + x^2} = 1 - \frac{\sin^2 x}{1 + x^2}$
Reciprocal Identity: $\csc^2 x \cdot \sin^2 x = 1$
Standard Integrals: $\int \csc^2 x \, dx = -\cot x + c$ and $\int \frac{1}{1 + x^2} \, dx = \tan^{-1} x + c$

$\int \left( \frac{x^2 + \cos^2 x}{1 + x^2} \right) \csc^2 x \, dx =$

A

$\cot x - \tan^{-1} x + c$

B

$\cot x + \tan^{-1} x + c$

C

$-\cot x + \tan^{-1} x + c$

D

$-\cot x - \tan^{-1} x + c$

Question 18: Palette 18/107

DPT MCQ

Algebraic Manipulation (Adding and Subtracting Constant): $\frac{x}{x + 3} = \frac{x + 3 - 3}{x + 3} = 1 - \frac{3}{x + 3}$
Standard Integration Formula: $\int \frac{1}{x + a} \, dx = \ln\vert{}x + a\vert{} + c$

$\int \frac{x}{x + 3} \, dx$

A

$3\ln(x + 3) - x + c$

B

$x + 3\ln(x + 3) + c$

C

$x - 3\ln(x + 3) + c$

D

None of these

Question 19: Palette 19/107

DPT MCQ

Binomial Series Expansion: $(1 - x)^{-2} = 1 + 2x + 3x^2 + 4x^3 + \dots$ for $\vert{}x\vert{} < 1$
Sum of Infinite Geometric Series: $S_\infty = \frac{a}{1 - r}$
Integration Formula: $\int (1 - x)^{-2} \, dx = (1 - x)^{-1} + c = \frac{1}{1 - x} + c$

Given, $\vert{}x\vert{} < 1$, find $\int (1 + 2x + 3x^2 + 4x^3 + \dots) \, dx$

A

$(1 + x)^{-1} + c$

B

$(1 - x)^{-1} + c$

C

$(1 - x)^{-1} - 1 + c$

D

None of these

Question 20: Palette 20/107

DPT MCQ

Trigonometric Half-Angle Identity: $\cos^2 x = \frac{1 + \cos 2x}{2}$
Double-Angle Identity: $2\sin 2x \cos 2x = \sin 4x$
Standard Integrals: $\int \sin(kx) \, dx = -\frac{\cos(kx)}{k} + c$

$\int \sin 2x \cos^2 x \, dx =$

A

$-\frac{\cos 2x}{2} - \frac{\cos 4x}{8} + c$

B

$\frac{\cos 2x}{4} + \frac{\cos 4x}{16} + c$

C

$-\frac{\cos 2x}{4} - \frac{\cos 4x}{16} + c$

D

None of these

Question 21: Palette 21/107

DPT MCQ

Trigonometric Double-Angle Identities: $\cos 4x + 1 = 2\cos^2 2x$ and $\cot x - \tan x = 2\cot 2x$
Reciprocal Identity: $\frac{\cos 2x}{\cot 2x} = \sin 2x$
Standard Integral: $\int \sin(kx) \, dx = -\frac{\cos(kx)}{k} + c$

$\int \frac{\cos 4x + 1}{\cot x - \tan x} \, dx =$

A

$-\frac{\cos 4x}{8} + c$

B

$\frac{\cos 4x}{8} + c$

C

$-\frac{\sin 4x}{8} + c$

D

None of these

Question 22: Palette 22/107

DPT MCQ

Algebraic Splitting: $\frac{2 + 3x^2}{x^2(1 + x^2)} = \frac{2(1 + x^2) + x^2}{x^2(1 + x^2)}$
Power Rule and Inverse Tangent Integration: $\int x^{-2} \, dx = \frac{x^{-1}}{-1} + c$ and $\int \frac{1}{1 + x^2} \, dx = \tan^{-1} x + c$

$\int \frac{2 + 3x^2}{x^2(1 + x^2)} \, dx =$

A

$-\frac{2}{x} - \tan^{-1} x + c$

B

$\frac{2}{x} + \tan^{-1} x + c$

C

$-\frac{2}{x} + \tan^{-1} x + c$

D

None of these

Question 23: Palette 23/107

DPT MCQ

Algebraic Splitting: $\frac{x^2(x^2 + 1) + 1}{2(1 + x^2)} = \frac{x^2}{2} + \frac{1}{2(1 + x^2)}$
Standard Integrals: $\int x^2 \, dx = \frac{x^3}{3} + c$ and $\int \frac{1}{1 + x^2} \, dx = \tan^{-1} x + c$

$\int \frac{(x^4 + x^2) + 1}{2(1 + x^2)} \, dx =$

A

$\frac{x^3}{3} + \tan^{-1} x + c$

B

$\frac{x^3}{6} + \frac{1}{2}\tan^{-1} x + c$

C

$\frac{x^3}{6} - \frac{1}{2}\tan^{-1} x + c$

D

None of these

Question 24: Palette 24/107

DPT MCQ

Algebraic Factoring: $x^2 - \sqrt{x} = \sqrt{x}(x^{\frac{3}{2}} - 1) = \sqrt{x}(\sqrt{x} - 1)(x + \sqrt{x} + 1)$
Denominator Factoring: $x\sqrt{x} + x + \sqrt{x} = \sqrt{x}(x + \sqrt{x} + 1)$
Difference of Squares: $(\sqrt{x} + 1)(\sqrt{x} - 1) = x - 1$
Power Rule of Integration: $\int x^n \, dx = \frac{x^{n+1}}{n+1} + c$

$\int \frac{(\sqrt{x} + 1)(x^2 - \sqrt{x})}{x\sqrt{x} + x + \sqrt{x}} \, dx =$

A

$\frac{x^2}{2} - x + c$

B

$\frac{x^2}{2} + x + c$

C

$x^2 - x + c$

D

None of these

Question 25: Palette 25/107

DPT MCQ

Algebraic Expansion: $(x + 1)^2 = x^2 + 2x + 1 = (x^2 + 1) + 2x$
Splitting Fractions: $\frac{(x^2 + 1) + 2x}{x(x^2 + 1)} = \frac{x^2 + 1}{x(x^2 + 1)} + \frac{2x}{x(x^2 + 1)} = \frac{1}{x} + \frac{2}{x^2 + 1}$
Standard Integrals: $\int \frac{1}{x} \, dx = \log_e x + c$ and $\int \frac{1}{x^2 + 1} \, dx = \tan^{-1} x + c$

$\int \frac{(x + 1)^2}{x(x^2 + 1)} \, dx$ is equal to [MP PET 2003]

A

$\log_e x + c$

B

$\log_e x + 2\tan^{-1} x + c$

C

$\log_e \frac{1}{x^2} + c$

D

$\log_e \{x(x^2 + 1)\} + c$

Question 26: Palette 26/107

DPT MCQ

Trigonometric Identity: $1 + \sin 2\theta = (\sin \theta + \cos \theta)^2$
Algebraic Difference of Squares: $a^2 - b^2 = (a - b)(a + b)$
Standard Integrals: $\int \sin x \, dx = -\cos x + c$ and $\int \cos x \, dx = \sin x + c$

$\int \frac{\sin 2x - \sin 2k}{\sin x - \sin k + \cos x - \cos k} \, dx =$

A

$\cos x + \sin x - x\sin k - x\cos k + c$

B

$-\cos x - \sin x + x\sin k + x\cos k + c$

C

$-\cos x + \sin x + x\sin k + x\cos k + c$

D

None of these

Question 27: Palette 27/107

DPT MCQ

Algebraic Splitting: $\frac{x^2 + 3}{x^6(x^2 + 1)} = \frac{(x^2 + 1) + 2}{x^6(x^2 + 1)} = \frac{1}{x^6} + \frac{2}{x^6(x^2 + 1)}$
Algebraic Trick for Partial Fractions: $\frac{1}{x^6(x^2 + 1)} = \frac{(1 + x^2) - x^2}{x^6(x^2 + 1)} = \frac{1}{x^6} - \frac{1}{x^4(x^2 + 1)}$
Power Rule and Inverse Tangent Integration: $\int x^{-n} \, dx = \frac{x^{-n+1}}{-n+1} + c$ and $\int \frac{1}{x^2 + 1} \, dx = \tan^{-1} x + c$

$\int \frac{x^2 + 3}{x^6(x^2 + 1)} \, dx =$

A

$-\frac{3}{5x^5} - \frac{2}{3x^3} + \frac{2}{x} + 2\tan^{-1} x + c$

B

$-\frac{3}{5x^5} + \frac{2}{3x^3} - \frac{2}{x} - 2\tan^{-1} x + c$

C

$\frac{3}{5x^5} - \frac{2}{3x^3} + \frac{2}{x} - 2\tan^{-1} x + c$

D

None of these

Question 28: Palette 28/107

DPT MCQ

Integration by Substitution: If an integrand is of the form $f(g(x))g'(x)$, we substitute $g(x) = t$, which gives $g'(x)dx = dt$, simplifying the integral to $\int f(t) dt$. The formula used is $\int \frac{1}{t} dt = \ln\vert{}t\vert{} + c$.

Evaluate the integral $\int \frac{x^3 dx}{x^4 + 1}$

A

$\frac{1}{2} \ln(1 + x^4) + c$

B

$\frac{1}{4} \ln(1 + x^4) + c$

C

$\ln(1 + x^4) + c$

D

$\frac{1}{4} \tan^{-1}(x^2) + c$

Question 29: Palette 29/107

DPT MCQ

Integration by Substitution: If an integral contains a function $g(x)$ and its derivative $g'(x)$, we substitute $g(x) = t$, giving $g'(x)dx = dt$. The integration formula used is $\int \frac{1}{t^2 + 1} dt = \tan^{-1}(t) + c$.

Evaluate the integral $\int \frac{xdx}{x^4 + 1}$

A

$\frac{1}{2} \tan^{-1}(x^2) + c$

B

$\frac{1}{4} \ln(1 + x^4) + c$

C

$\tan^{-1}(x^2) + c$

D

$\frac{1}{2} \ln(1 + x^2) + c$

Question 30: Palette 30/107

DPT MCQ

Integration by Substitution: When an integrand contains a inverse trigonometric function alongside its derivative, we substitute $\sin^{-1}x = t$, giving $\frac{1}{\sqrt{1 - x^2}}dx = dt$. The integration formula used is $\int t^n dt = \frac{t^{n+1}}{n+1} + c$.

Evaluate the integral $\int \sin^{-1}x \frac{dx}{\left(\sqrt{1 - x^2}\right)}$

A

$\frac{(\sin^{-1}x)^2}{2} + c$

B

$\sin^{-1}x + c$

C

$\frac{1}{2\sqrt{1 - x^2}} + c$

D

$(\sin^{-1}x)^2 + c$

Question 31: Palette 31/107

DPT MCQ

Integration by Substitution: When an integrand contains a composite function with an inner function $g(x)$ whose derivative $g'(x)$ is also present in the integral, we substitute $g(x) = \theta$, giving $g'(x)dx = d\theta$. The standard formula used is $\int \sin \theta \, d\theta = -\cos \theta + c$.

Evaluate the integral $\int x^4 \sin(x^5) dx$

A

$-\frac{1}{5} \cos(x^5) + c$

B

$\frac{1}{5} \cos(x^5) + c$

C

$-\cos(x^5) + c$

D

$-\frac{1}{4} \cos(x^5) + c$

Question 32: Palette 32/107

DPT MCQ

Integration by Substitution: If an integrand contains an exponential term $a^x$ and its derivative $a^x \ln a$, we substitute $a^x = t$, giving $a^x \ln a \cdot dx = dt$. The standard formula used is $\int \frac{1}{\sqrt{1 - t^2}} dt = \sin^{-1}(t) + c$.

Evaluate the integral $\int \frac{a^x}{\sqrt{1 - a^{2x}}} dx$

A

$\frac{1}{\log a} \sin^{-1}(a^x) + c$

B

$\sin^{-1}(a^x) + c$

C

$\frac{1}{\log a} \cos^{-1}(a^x) + c$

D

$\cos^{-1}(a^x) + c$

Question 33: Palette 33/107

DPT MCQ

Exponential and Logarithmic Property with Integration by Substitution: The logarithmic identity $e^{\log f(x)} = f(x)$ simplifies the integrand to $\cos^3 x \sin x$. We then use substitution by letting $\cos x = t$, which gives $-\sin x dx = dt$. The power rule of integration $\int t^n dt = \frac{t^{n+1}}{n+1} + c$ is then applied.

Evaluate the integral $\int (\cos^3 x) e^{\log(\sin x)} dx$

A

$-\frac{\sin^4 x}{4} + c$

B

$-\frac{\cos^4 x}{4} + c$

C

$\frac{e^{\sin x}}{4} + c$

D

None of these

Question 34: Palette 34/107

DPT MCQ

Integration by Substitution: Using the trigonometric identity $\frac{1}{\sin^2 x} = \csc^2 x$, we rewrite the integral as $\int \frac{6\csc^2 x}{(1 - \cot x)^2} dx$. Then, substituting $1 - \cot x = t$ gives $\csc^2 x dx = dt$. The power rule $\int t^{-2} dt = -\frac{1}{t} + c$ is used to integrate, followed by evaluating the constant of integration $c$ and the specific function value $I\left(\frac{\pi}{12}\right)$ using $\cot\left(\frac{\pi}{12}\right) = 2 + \sqrt{3}$.

Let $I(x) = \int \frac{6}{\sin^2 x(1 - \cot x)^2} dx$. If $I(0) = 3$, then $I\left(\frac{\pi}{12}\right)$ is equal to

A

$2\sqrt{3}$

B

$\sqrt{3}$

C

$3\sqrt{3}$

D

$6\sqrt{3}$

Question 35: Palette 35/107

DPT MCQ

Integration by Substitution: Divide the numerator and denominator by $\cos^2 x$ to rewrite the integrand in terms of $\tan x$ and $\sec^2 x$. Then, substitute $\tan x = t^2$, which gives $\sec^2 x dx = 2t dt$. The formula used is $\int 1 dt = t + c$.

Evaluate the integral $\int \frac{\sqrt{\tan x}}{\sin x \cos x} dx$

A

$2\sqrt{\sec x} + c$

B

$2\sqrt{\tan x} + c$

C

$\frac{2}{\sqrt{\tan x}} + c$

D

$\frac{2}{\sqrt{\sec x}} + c$

Question 36: Palette 36/107

DPT MCQ

Integration by Substitution Method:
If $t = g(x)$, then $dt = g'(x) dx$.
Standard Integral Formula:
$\int t^n dt = \frac{t^{n+1}}{n+1} + c$

$\int (\frac{1}{\sqrt{x}}) \tan^4 \sqrt{x} (\sec^2 \sqrt{x}) dx$

A

$2 \tan^5 \sqrt{x} + c$

B

$\frac{1}{5} \tan^5 \sqrt{x} + c$

C

$\frac{2}{5} \tan^5 \sqrt{x} + c$

D

None of these

Question 37: Palette 37/107

DPT MCQ

Integration by Substitution Method:
Trigonometric identity: $\cos^2 x = 1 - \sin^2 x$
Substituted derivative: If $t = \sin x$, then $dt = \cos x dx$.
Standard Integral Formula:
$\int t^n dt = \frac{t^{n+1}}{n+1} + c$

$\int \cos^3 x dx$

A

$\sin x + \frac{\sin^3 x}{3} + c$

B

$\sin x - \frac{\sin^3 x}{3} + c$

C

$\cos x - \frac{\cos^3 x}{3} + c$

D

None of these

Question 38: Palette 38/107

DPT MCQ

Integration by Substitution Method:
Trigonometric identity: $\cos^2 x = 1 - \sin^2 x \Rightarrow \cos^4 x = (1 - \sin^2 x)^2$
Substituted derivative: If $t = \sin x$, then $dt = \cos x dx$.
Standard Integral Formula:
$\int t^n dt = \frac{t^{n+1}}{n+1} + c$

$\int \cos^5 x dx$

A

$\sin x + \frac{\sin^5 x}{5} - \frac{2\sin^3 x}{3} + c$

B

$\sin x - \frac{\sin^5 x}{5} + \frac{2\sin^3 x}{3} + c$

C

$\cos x + \frac{\cos^5 x}{5} - \frac{2\cos^3 x}{3} + c$

D

None of these

Question 39: Palette 39/107

DPT MCQ

Integration using Power Reduction Formulas (Half-Angle Formulas):
$\cos^2 x = \frac{1 + \cos 2x}{2}$
$\cos^2 2x = \frac{1 + \cos 4x}{2}$
Standard Integral Formulas:
$\int \cos(ax) dx = \frac{\sin(ax)}{a} + c$
$\int 1 dx = x + c$

$\int \cos^4 x dx$

A

$\frac{3x}{8} + \frac{\sin 2x}{4} + \frac{\sin 4x}{32} + c$

B

$\frac{3x}{8} - \frac{\sin 2x}{4} + \frac{\sin 4x}{32} + c$

C

$\frac{x}{8} + \frac{\sin 2x}{4} + \frac{\sin 4x}{16} + c$

D

None of these

Question 40: Palette 40/107

DPT MCQ

Integration by Substitution Method:
If $x^3 = t$, then $3x^2 dx = dt \Rightarrow x^2 dx = \frac{dt}{3}$.
Standard Integral Formula:
$\int \sec t dt = \ln(\sec t + \tan t) + c = \log(\sec t + \tan t) + c$

$\int x^2 \sec x^3 dx =$

A

$\log (\sec x^3 + \tan x^3)$

B

$3(\sec x^3 + \tan x^3)$

C

$\frac{1}{3}\log (\sec x^3 + \tan x^3)$

D

None of these

Question 41: Palette 41/107

DPT MCQ

Integration by Substitution Method:
Substitute $3 + 5x^4 = t^2$, then $20x^3 dx = 2t dt \Rightarrow x^3 dx = \frac{tdt}{10}$.
Standard Integral Formula:
$\int t^2 dt = \frac{t^3}{3} + c$

$\int x^3 \sqrt{3 + 5x^4} dx =$

A

$(3 + 5x^4)^{\frac{3}{2}} + c$

B

$\frac{1}{5}(3 + 5x^4)^{\frac{3}{2}} + c$

C

$\frac{1}{30}(3 + 5x^4)^{\frac{3}{2}} + c$

D

None of these

Question 42: Palette 42/107

DPT MCQ

Integration by Substitution Method:
Trigonometric identity: $1 + \sin 2x = (\cos x + \sin x)^2$
Substituted derivative: If $t = \cos x + \sin x$, then $dt = (\cos x - \sin x) dx$.
Standard Integral Formula:
$\int t^{-2} dt = \frac{t^{-2+1}}{-2+1} + c = -\frac{1}{t} + c$

$\int \frac{\cos x - \sin x}{1 + \sin 2x} dx =$

A

$-\frac{1}{\cos x + \sin x} + c$

B

$\frac{1}{\cos x + \sin x} + c$

C

$\frac{1}{\cos x - \sin x} + c$

D

None of these

Question 43: Palette 43/107

DPT MCQ

Integration by Substitution Method:
Substitute $e^{\sqrt{x}} = t$.
Using chain rule, $d(e^{\sqrt{x}}) = e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}} dx = dt \Rightarrow \frac{e^{\sqrt{x}}}{\sqrt{x}} dx = 2dt$.
Standard Integral Formula:
$\int \cos t dt = \sin t + c$

$\int \frac{e^{\sqrt{x}} \cos e^{\sqrt{x}}}{\sqrt{x}} dx =$

A

$2 \sin e^{\sqrt{x}} + c$

B

$\sin e^{\sqrt{x}} + c$

C

$2 \cos e^{\sqrt{x}} + c$

D

$-2 \sin e^{\sqrt{x}} + c$

Question 44: Palette 44/107

DPT MCQ

Integration using Constant Angle Transformation:
Multiply and divide by $\sin(a - b) = \sin((x - b) - (x - a))$.
Trigonometric Identity:
$\sin(A - B) = \sin A \cos B - \cos A \sin B$
Standard Integral Formula:
$\int \tan(kx) dx = \ln\vert{}\sec(kx)\vert{} + c$

Question 1:
$\int \frac{\tan(\sin^{-1} x)}{\sqrt{1 - x^2}} dx$

A

$\frac{1}{\sin(a - b)} \ln\left\vert{} \frac{\sec(x - a)}{\sec(x - b)} \right\vert{} + c$

B

$\frac{1}{\sin(b - a)} \ln\left\vert{} \frac{\cos(x - a)}{\cos(x - b)} \right\vert{} + c$

C

$\frac{1}{\cos(a - b)} \ln\left\vert{} \frac{\sec(x - a)}{\sec(x - b)} \right\vert{} + c$

D

None of these

Question 45: Palette 45/107

DPT MCQ

Integration by Algebraic Manipulation of Angle:
Rewrite $x$ in the numerator as $(x - a) + a$.
Trigonometric Addition Identity:
$\cos(A + B) = \cos A \cos B - \sin A \sin B$
Standard Integral Formulas:
$\int 1 dx = x + c$
$\int \tan(x - a) dx = \ln\vert{}\sec(x - a)\vert{} + c$

$\int \frac{\cos x}{\cos(x - a)} dx$

A

$x \sin a - \cos a \ln\vert{}\sec(x - a)\vert{} + c$

B

$x \cos a + \sin a \ln\vert{}\sec(x - a)\vert{} + c$

C

$x \cos a - \sin a \ln\vert{}\sec(x - a)\vert{} + c$

D

None of these

Question 46: Palette 46/107

DPT MCQ

Trigonometric Angle Decomposition:
Express $2x$ in the numerator as $(5x - 3x)$.
Trigonometric Sine Difference Identity:
$\sin(A - B) = \sin A \cos B - \cos A \sin B$
Standard Integral Formula:
$\int \cot(kx) dx = \frac{1}{k}\ln\vert{}\sin(kx)\vert{} + c$

$\int \frac{\sin 2x}{\sin 5x \sin 3x} dx$

A

$\frac{1}{5}\ln\vert{}\sin 5x\vert{} - \frac{1}{3}\ln\vert{}\sin 3x\vert{} + c$

B

$\frac{1}{3}\ln\vert{}\sin 3x\vert{} - \frac{1}{5}\ln\vert{}\sin 5x\vert{} + c$

C

$\frac{1}{2}\ln\vert{}\sin 2x\vert{} + c$

D

None of these

Question 47: Palette 47/107

DPT MCQ

Integration using Constant Angle Transformation:
Multiply and divide by $\sin(a - b) = \sin((x - b) - (x - a))$.
Trigonometric Identity:
$\sin(A - B) = \sin A \cos B - \cos A \sin B$
Standard Integral Formula:
$\int \tan(kx) dx = \ln\vert{}\sec(kx)\vert{} + c$

$\int \frac{dx}{\cos(x - a)\cos(x - b)}$

A

$\frac{1}{\sin(a - b)} \ln\left\vert{} \frac{\sec(x - a)}{\sec(x - b)} \right\vert{} + c$

B

$\frac{1}{\sin(b - a)} \ln\left\vert{} \frac{\cos(x - a)}{\cos(x - b)} \right\vert{} + c$

C

$\frac{1}{\cos(a - b)} \ln\left\vert{} \frac{\sec(x - a)}{\sec(x - b)} \right\vert{} + c$

D

None of these

Question 48: Palette 48/107

DPT MCQ

Constant multiple rule combined with the logarithmic integration rule from the image: $\int \frac{f'(x)}{f(x)} dx = \ln f(x) + C$, where the numerator is a constant multiple of the derivative of the denominator.

Evaluate $\int \frac{6x}{x^2 + 4} dx$.

A

$\ln(x^2 + 4) + C$

B

$6 \ln(x^2 + 4) + C$

C

$3 \ln(x^2 + 4) + C$

D

$\tan^{-1}(x/2) + C$

Question 49: Palette 49/107

DPT MCQ

Reverse power rule (generalised substitution) from the image: $\int f'(x) f^n(x) dx = \frac{f^{n+1}(x)}{n+1} + C$, with $f(x) = x^2 + 1$ and $n = 3$.

Evaluate $\int 2x (x^2 + 1)^3 dx$.

A

$\frac{(x^2 + 1)^4}{4} + C$

B

$\frac{(x^2 + 1)^3}{3} + C$

C

$\frac{(x^2 + 1)^2}{2} + C$

D

$(x^2 + 1)^4 + C$

Question 50: Palette 50/107

DPT MCQ

Exponential reverse-chain rule from the image: $\int f'(x) e^{f(x)} dx = e^{f(x)} + C$, with $f(x) = x^2$ and $f'(x) = 2x$.

Evaluate $\int 2x e^{x^2} dx$.

A

$e^{x^2} + C$

B

$\frac{e^{x^2}}{2} + C$

C

$x^2 e^{x^2} + C$

D

$e^{2x} + C$

Question 51: Palette 51/107

DPT MCQ

Square-root reverse-chain rule from the image: $\int \frac{f'(x)}{2\sqrt{f(x)}} dx = \sqrt{f(x)} + C$, with $f(x) = x^2 + 9$ and $f'(x) = 2x$ (note $\frac{2x}{2\sqrt{f}} = \frac{x}{\sqrt{f}}$).

Evaluate $\int \frac{x}{\sqrt{x^2 + 9}} dx$.

A

$\sqrt{x^2 + 9} + C$

B

$2\sqrt{x^2 + 9} + C$

C

$\frac{\sqrt{x^2 + 9}}{2} + C$

D

$\frac{(x^2 + 9)^{3/2}}{3} + C$

Question 52: Palette 52/107

DPT MCQ

Concept

+

Integral reducible to the inverse-tangent standard form; formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C$; rewrite $bx^2 + c$ as $(\sqrt{b}x)^2 + (\sqrt{c})^2$ and use substitution to match $\int \frac{dx}{x^2 + a^2}$

Evaluate $\int \frac{dx}{2x^2 + 7}$

A

$\frac{1}{\sqrt{2}} \frac{1}{\sqrt{7}} \tan^{-1} \left( \frac{\sqrt{2}x}{\sqrt{7}} \right) + C$

B

$\frac{1}{\sqrt{2}} \frac{1}{\sqrt{7}} \tan^{-1} \left( \frac{\sqrt{7}x}{\sqrt{2}} \right) + C$

C

$\sqrt{2} \sqrt{7} \tan^{-1} \left( \frac{\sqrt{2}x}{\sqrt{7}} \right) + C$

D

$\frac{1}{2} \frac{1}{7} \tan^{-1} \left( \frac{2x}{7} \right) + C$

Question 53: Palette 53/107

DPT MCQ

Concept

+

Integration by partial fractions leading to the logarithmic standard form; formula $\int \frac{dx}{x^2 - a^2} = \frac{1}{2a} \ln \left| \frac{x-a}{x+a} \right| + C$; identify $a$ from $x^2 - a^2$ and substitute directly

Evaluate $\int \frac{dx}{x^2 - 16}$

A

$\frac{1}{4} \ln \left| \frac{x-4}{x+4} \right| + C$

B

$\frac{1}{8} \ln \left| \frac{x-4}{x+4} \right| + C$

C

$\frac{1}{8} \ln \left| \frac{x+4}{x-4} \right| + C$

D

$\frac{1}{4} \tan^{-1} \left( \frac{x}{4} \right) + C$

Question 54: Palette 54/107

DPT MCQ

Concept

+

Integration by partial fractions leading to the logarithmic standard form; formula $\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \ln \left| \frac{a+x}{a-x} \right| + C$; identify $a$ from $a^2 - x^2$ and substitute directly

Evaluate $\int \frac{dx}{9 - x^2}$

A

$\frac{1}{3} \ln \left| \frac{3+x}{3-x} \right| + C$

B

$\frac{1}{6} \ln \left| \frac{3-x}{3+x} \right| + C$

C

$\frac{1}{6} \ln \left| \frac{3+x}{3-x} \right| + C$

D

$\frac{1}{6} \tan^{-1} \left( \frac{x}{3} \right) + C$

Question 55: Palette 55/107

DPT MCQ

Concept

+

Standard inverse-sine integral; formula $\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1} \left( \frac{x}{a} \right) + C$; rewrite the constant under the root as $a^2$ and substitute directly

Evaluate $\int \frac{dx}{\sqrt{9 - x^2}}$

A

$\sin^{-1} \left( \frac{x}{9} \right) + C$

B

$\sin^{-1} \left( \frac{x}{3} \right) + C$

C

$\cos^{-1} \left( \frac{x}{3} \right) + C$

D

$\frac{1}{3} \sin^{-1} \left( \frac{x}{3} \right) + C$

Question 56: Palette 56/107

DPT MCQ

Concept

+

Standard logarithmic-form integral for a sum-of-squares root; formula $\int \frac{dx}{\sqrt{x^2 + a^2}} = \ln \left| x + \sqrt{x^2 + a^2} \right| + C$; rewrite the constant as $a^2$ and substitute directly

Evaluate $\int \frac{dx}{\sqrt{x^2 + 9}}$

A

$\ln \left| x + \sqrt{x^2 + 9} \right| + C$

B

$\ln \left| x + \sqrt{x^2 + 3} \right| + C$

C

$\frac{1}{3} \ln \left| x + \sqrt{x^2 + 9} \right| + C$

D

$\sin^{-1} \left( \frac{x}{3} \right) + C$

Question 57: Palette 57/107

DPT MCQ

Concept

+

Standard logarithmic-form integral for a difference-of-squares root; formula $\int \frac{dx}{\sqrt{x^2 - a^2}} = \ln \left| x + \sqrt{x^2 - a^2} \right| + C$; rewrite the constant as $a^2$ and substitute directly

Evaluate $\int \frac{dx}{\sqrt{x^2 - 9}}$

A

$\ln \left| x - \sqrt{x^2 - 9} \right| + C$

B

$\frac{1}{3} \ln \left| x + \sqrt{x^2 - 9} \right| + C$

C

$\ln \left| x + \sqrt{x^2 - 9} \right| + C$

D

$\sin^{-1} \left( \frac{x}{3} \right) + C$

Question 58: Palette 58/107

DPT MCQ

Concept

+

Integration of a difference-of-squares root (via substitution $x = a \sin \theta$ or the standard result); formula $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1} \left( \frac{x}{a} \right) + C$; rewrite the constant as $a^2$ and substitute directly

Evaluate $\int \sqrt{9 - x^2} \, dx$

A

$\frac{x}{2} \sqrt{9 - x^2} + \frac{9}{2} \sin^{-1} \left( \frac{x}{9} \right) + C$

B

$\frac{x}{2} \sqrt{9 - x^2} + \frac{9}{2} \sin^{-1} \left( \frac{x}{3} \right) + C$

C

$\frac{x}{2} \sqrt{9 - x^2} - \frac{9}{2} \sin^{-1} \left( \frac{x}{3} \right) + C$

D

$\frac{x}{2} \sqrt{9 - x^2} + \frac{3}{2} \sin^{-1} \left( \frac{x}{3} \right) + C$

Question 59: Palette 59/107

DPT MCQ

Concept

+

Integration of a sum-of-squares root (via substitution $x = a \tan \theta$ or the standard result); formula $\int \sqrt{x^2 + a^2} \, dx = \frac{x}{2} \sqrt{x^2 + a^2} + \frac{a^2}{2} \ln \left| x + \sqrt{x^2 + a^2} \right| + C$; rewrite the constant as $a^2$ and substitute directly

Evaluate $\int \sqrt{x^2 + 9} \, dx$

A

$\frac{x}{2} \sqrt{x^2 + 9} + \frac{9}{2} \sin^{-1} \left( \frac{x}{3} \right) + C$

B

$\frac{x}{2} \sqrt{x^2 + 9} - \frac{9}{2} \ln \left| x + \sqrt{x^2 + 9} \right| + C$

C

$\frac{x}{2} \sqrt{x^2 + 9} + \frac{9}{2} \ln \left| x + \sqrt{x^2 + 9} \right| + C$

D

$\frac{x}{2} \sqrt{x^2 + 9} + \frac{3}{2} \ln \left| x + \sqrt{x^2 + 9} \right| + C$

Question 60: Palette 60/107

DPT MCQ

Concept

+

Integration of a difference-of-squares root (via substitution $x = a \sec \theta$ or the standard result); formula $\int \sqrt{x^2 - a^2} \, dx = \frac{x}{2} \sqrt{x^2 - a^2} - \frac{a^2}{2} \ln \left| x + \sqrt{x^2 - a^2} \right| + C$; rewrite the constant as $a^2$ and substitute directly

Evaluate $\int \sqrt{x^2 - 9} \, dx$

A

$\frac{x}{2} \sqrt{x^2 - 9} + \frac{9}{2} \ln \left| x + \sqrt{x^2 - 9} \right| + C$

B

$\frac{x}{2} \sqrt{x^2 - 9} - \frac{9}{2} \sin^{-1} \left( \frac{x}{3} \right) + C$

C

$\frac{x}{2} \sqrt{x^2 - 9} - \frac{3}{2} \ln \left| x + \sqrt{x^2 - 9} \right| + C$

D

$\frac{x}{2} \sqrt{x^2 - 9} - \frac{9}{2} \ln \left| x + \sqrt{x^2 - 9} \right| + C$

Question 61: Palette 61/107

DPT MCQ

Concept

+

Integration by partial fractions: factorise the quadratic denominator into linear factors, split using $\frac{1}{(x-a)(x-b)} = \frac{1}{a-b}\left(\frac{1}{x-a} - \frac{1}{x-b}\right)$, and integrate using $\int \frac{dx}{x-a} = \ln|x-a| + C$

Evaluate $\int \frac{dx}{x^2 - x - 6}$

A

$\frac{1}{5} \ln \left| \frac{x+2}{x-3} \right| + C$

B

$\frac{1}{5} \ln \left| \frac{x-3}{x+2} \right| + C$

C

$\frac{1}{6} \ln \left| \frac{x-3}{x+2} \right| + C$

D

$\ln \left| \frac{x-3}{x+2} \right| + C$

Question 62: Palette 62/107

DPT MCQ

Concept

+

Integration by partial fractions: factorise the quadratic denominator into distinct linear factors, split using $\frac{1}{(x-a)(x-b)} = \frac{1}{a-b}\left(\frac{1}{x-a} - \frac{1}{x-b}\right)$, and integrate using $\int \frac{dx}{x-a} = \ln|x-a| + C$

Evaluate $\int \frac{dx}{x^2 - 7x + 10}$

A

$\frac{1}{3} \ln \left| \frac{x-2}{x-5} \right| + C$

B

$\frac{1}{3} \ln \left| \frac{x-5}{x-2} \right| + C$

C

$\ln \left| \frac{x-5}{x-2} \right| + C$

D

$\frac{1}{2} \ln \left| \frac{x-5}{x-2} \right| + C$

Question 63: Palette 63/107

DPT MCQ

Concept

+

Substitution reducing to a rational integral: put $t = \sin^2 x$ so that $dt = 2 \sin x \cos x \, dx = \sin 2x \, dx$, then partial fractions with $\int \frac{dt}{(t-a)(t-b)} = \frac{1}{a-b} \ln \left| \frac{t-a}{t-b} \right| + C$

Evaluate $\int \frac{\sin 2x \, dx}{(\sin^2 x - 1)(\sin^2 x + 3)}$

A

$\frac{1}{4} \ln \left| \frac{\sin^2 x + 3}{\sin^2 x - 1} \right| + C$

B

$\frac{1}{4} \ln \left| \frac{\sin^2 x - 1}{\sin^2 x + 3} \right| + C$

C

$\frac{1}{2} \ln \left| \frac{\sin^2 x - 1}{\sin^2 x + 3} \right| + C$

D

$\ln \left| \frac{\sin^2 x - 1}{\sin^2 x + 3} \right| + C$

Question 64: Palette 64/107

DPT MCQ

Concept

+

Substitution reducing to a rational integral: put $t = e^{2x}$ so that $dt = 2e^{2x} \, dx$, then partial fractions with $\int \frac{dt}{(t-a)(t-b)} = \frac{1}{a-b} \ln \left| \frac{t-a}{t-b} \right| + C$

Evaluate $\int \frac{2e^{2x} \, dx}{(e^{2x} - 1)(e^{2x} - 9)}$

A

$\frac{1}{8} \ln \left| \frac{e^{2x} - 1}{e^{2x} - 9} \right| + C$

B

$\frac{1}{4} \ln \left| \frac{e^{2x} - 9}{e^{2x} - 1} \right| + C$

C

$\frac{1}{8} \ln \left| \frac{e^{2x} - 9}{e^{2x} - 1} \right| + C$

D

$\ln \left| \frac{e^{2x} - 9}{e^{2x} - 1} \right| + C$

Question 65: Palette 65/107

DPT MCQ

Concept

+

Completing the square to obtain the standard form $\int \frac{du}{u^2 + a^2} = \frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right) + C$; here $x^2 - 2x + 3 = (x-1)^2 + 2$ with $u = x-1$ and $a = \sqrt{2}$

Evaluate $\int \frac{dx}{x^2 - 2x + 3}$

A

$\frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{x+1}{\sqrt{2}}\right) + C$

B

$\frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{x-1}{\sqrt{2}}\right) + C$

C

$\frac{1}{2} \tan^{-1}\left(\frac{x-1}{2}\right) + C$

D

$\sqrt{2} \tan^{-1}\left(\frac{x-1}{\sqrt{2}}\right) + C$

Question 66: Palette 66/107

DPT MCQ

Concept

+

Complete the square to obtain the form $\sqrt{a^2 - u^2}$: here $3x - x^2 = \left(\frac{3}{2}\right)^2 - \left(x - \frac{3}{2}\right)^2$, then use $\int \sqrt{a^2 - u^2} \, du = \frac{u}{2}\sqrt{a^2 - u^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{u}{a}\right) + C$

Evaluate $\int \sqrt{3x - x^2} \, dx$

A

$\frac{2x - 3}{2} \sqrt{3x - x^2} + \frac{9}{4} \sin^{-1}\left(\frac{2x - 3}{3}\right) + C$

B

$\frac{2x - 3}{4} \sqrt{3x - x^2} - \frac{9}{8} \sin^{-1}\left(\frac{2x - 3}{3}\right) + C$

C

$\frac{2x - 3}{4} \sqrt{3x - x^2} + \frac{9}{8} \sin^{-1}\left(\frac{2x - 3}{3}\right) + C$

D

$\frac{2x + 3}{4} \sqrt{3x - x^2} + \frac{9}{8} \sin^{-1}\left(\frac{2x + 3}{3}\right) + C$

Question 67: Palette 67/107

DPT MCQ

Concept

+

Make the leading coefficient 1, complete the square to obtain the form $\sqrt{u^2 - a^2}$, and use $\int \frac{du}{\sqrt{u^2 - a^2}} = \ln\left| u + \sqrt{u^2 - a^2} \right| + C$

Evaluate $\int \frac{dx}{\sqrt{3x^2 + x - 2}}$

A

$\sqrt{3} \ln\left| 6x + 1 + 2\sqrt{3} \sqrt{3x^2 + x - 2} \right| + C$

B

$\frac{1}{\sqrt{3}} \ln\left| 6x - 1 + 2\sqrt{3} \sqrt{3x^2 + x - 2} \right| + C$

C

$\frac{1}{\sqrt{3}} \ln\left| 6x + 1 + 2\sqrt{3} \sqrt{3x^2 + x - 2} \right| + C$

D

$\frac{1}{\sqrt{3}} \sin^{-1}\left(\frac{6x + 1}{5}\right) + C$

Question 68: Palette 68/107

DPT MCQ

Concept

+

Make the leading coefficient 1 and complete the square; since the discriminant is negative the quadratic is irreducible, giving the form $u^2 + a^2$, and use $\int \frac{du}{u^2 + a^2} = \frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right) + C$

Evaluate $\int \frac{dx}{3x^2 + 7x + 6}$

A

$\frac{2}{\sqrt{23}} \tan^{-1}\left(\frac{6x - 7}{\sqrt{23}}\right) + C$

B

$\frac{1}{\sqrt{23}} \tan^{-1}\left(\frac{6x + 7}{\sqrt{23}}\right) + C$

C

$\frac{2}{\sqrt{23}} \tan^{-1}\left(\frac{6x + 7}{\sqrt{23}}\right) + C$

D

$\frac{2}{\sqrt{23}} \tan^{-1}\left(\frac{6x + 7}{23}\right) + C$

Question 69: Palette 69/107

DPT MCQ

Concept

+

Integration of a rational function with linear numerator and linear denominator; express the numerator as a multiple of the denominator plus a constant, then use $\int \frac{1}{ax+b} \, dx = \frac{1}{a}\ln|ax+b| + C$ and the general formula $\int \frac{ax+b}{cx+d} \, dx = \frac{a}{c}x + \frac{bc-ad}{c^2}\ln|cx+d| + C$

Evaluate $\int \frac{2x+7}{3x+4} \, dx$

A

$\frac{2}{3}x + \frac{13}{3}\ln|3x+4| + C$

B

$\frac{2}{3}x + \frac{13}{9}\ln|3x+4| + C$

C

$\frac{3}{2}x + \frac{13}{9}\ln|3x+4| + C$

D

$\frac{2}{3}x - \frac{13}{9}\ln|3x+4| + C$

Question 70: Palette 70/107

DPT MCQ

Concept

+

Integration of a rational function with linear numerator and linear denominator; express the numerator as a multiple of the denominator plus a constant, then use $\int \frac{1}{ax+b} \, dx = \frac{1}{a}\ln|ax+b| + C$ and the general formula $\int \frac{ax+b}{cx+d} \, dx = \frac{a}{c}x + \frac{bc-ad}{c^2}\ln|cx+d| + C$

Evaluate $\int \frac{3x-4}{5x+3} \, dx$

A

$\frac{3}{5}x + \frac{29}{25}\ln|5x+3| + C$

B

$\frac{5}{3}x - \frac{29}{25}\ln|5x+3| + C$

C

$\frac{3}{5}x - \frac{29}{25}\ln|5x+3| + C$

D

$\frac{3}{5}x - \frac{29}{5}\ln|5x+3| + C$

Question 71: Palette 71/107

DPT MCQ

Concept

+

Integration of a rational function with linear numerator and linear denominator; express the numerator as a multiple of the denominator plus a constant, then use $\int \frac{1}{ax+b} \, dx = \frac{1}{a}\ln|ax+b| + C$ and the general formula $\int \frac{ax+b}{cx+d} \, dx = \frac{a}{c}x + \frac{bc-ad}{c^2}\ln|cx+d| + C$

Evaluate $\int \frac{x}{3x+7} \, dx$

A

$\frac{1}{3}x + \frac{7}{9}\ln|3x+7| + C$

B

$\frac{1}{3}x - \frac{7}{9}\ln|3x+7| + C$

C

$\frac{1}{3}x - \frac{7}{3}\ln|3x+7| + C$

D

$3x - \frac{7}{9}\ln|3x+7| + C$

Question 72: Palette 72/107

DPT MCQ

Concept

+

Integration of an irrational function of the form $\frac{x}{\sqrt{ax+b}}$; express the numerator as a multiple of the linear expression under the root plus a constant, then use $\int (ax+b)^n \, dx = \frac{(ax+b)^{n+1}}{a(n+1)} + C$ and $\int \frac{dx}{\sqrt{ax+b}} = \frac{2\sqrt{ax+b}}{a} + C$

Evaluate $\int \frac{x}{\sqrt{2x+1}} \, dx$

A

$\frac{1}{6}(2x+1)^{3/2} + \frac{1}{2}\sqrt{2x+1} + C$

B

$\frac{1}{6}(2x+1)^{3/2} - \frac{1}{2}\sqrt{2x+1} + C$

C

$\frac{1}{3}(2x+1)^{3/2} - \frac{1}{2}\sqrt{2x+1} + C$

D

$\frac{1}{6}(2x+1)^{3/2} - \sqrt{2x+1} + C$

Question 73: Palette 73/107

DPT MCQ

Concept

+

Integration of a rational function (linear over quadratic): express the numerator as $\frac{1}{2}$(derivative of denominator) plus a constant; use $\int \frac{f'(x)}{f(x)} \, dx = \ln|f(x)| + C$, complete the square, and apply $\int \frac{dX}{X^2-a^2} = \frac{1}{2a}\ln\left|\frac{X-a}{X+a}\right| + C$

Evaluate $\int \frac{x+2}{x^2-3x+1} \, dx$

A

$\frac{1}{2}\ln|x^2-3x+1| - \frac{7}{2\sqrt{5}}\ln\left|\frac{x-\frac{3}{2}-\frac{\sqrt{5}}{2}}{x-\frac{3}{2}+\frac{\sqrt{5}}{2}}\right| + C$

B

$\ln|x^2-3x+1| + \frac{7}{2\sqrt{5}}\ln\left|\frac{x-\frac{3}{2}-\frac{\sqrt{5}}{2}}{x-\frac{3}{2}+\frac{\sqrt{5}}{2}}\right| + C$

C

$\frac{1}{2}\ln|x^2-3x+1| + \frac{7}{2\sqrt{5}}\ln\left|\frac{x-\frac{3}{2}-\frac{\sqrt{5}}{2}}{x-\frac{3}{2}+\frac{\sqrt{5}}{2}}\right| + C$

D

$\frac{1}{2}\ln|x^2-3x+1| + \frac{7}{\sqrt{5}}\ln\left|\frac{x-\frac{3}{2}-\frac{\sqrt{5}}{2}}{x-\frac{3}{2}+\frac{\sqrt{5}}{2}}\right| + C$

Question 74: Palette 74/107

DPT MCQ

Concept

+

Integral of the type $\int \frac{f'(x)}{f(x)} \, dx = \log|f(x)| + C$; here the numerator is half the derivative of the denominator, so the result is $\frac{1}{2}\log|f(x)| = \log\sqrt{f(x)} + C$

Evaluate $\int \frac{x-2}{x^2-4x+3} \, dx$

A

$\log\sqrt{x^2-4x+3} + C$

B

$x\log(x-3) - 2\log(x-2) + C$

C

$\log[(x-3)(x-1)] + C$

D

None of these

Question 75: Palette 75/107

DPT MCQ

Concept

+

Integral of type $\int \frac{px+q}{\sqrt{ax^2+bx+c}} \, dx$: express $px+q = A\frac{d}{dx}(ax^2+bx+c) + B$; the first part gives $2A\sqrt{ax^2+bx+c}$ via substitution $t = \sqrt{ax^2+bx+c}$, and the second part uses completing the square with $\int \frac{du}{\sqrt{u^2+k^2}} = \log|u+\sqrt{u^2+k^2}| + C$

Evaluate $\int \frac{2x-3}{\sqrt{4x^2+4x+2}} \, dx$

A

$\frac{1}{2}\sqrt{4x^2+4x+2} + 2\log|(2x+1)+\sqrt{4x^2+4x+2}| + C$

B

$\frac{1}{2}\sqrt{4x^2+4x+2} - 2\log|(2x+1)+\sqrt{4x^2+4x+2}| + C$

C

$\sqrt{4x^2+4x+2} - 2\log|(2x+1)+\sqrt{4x^2+4x+2}| + C$

D

$\frac{1}{2}\sqrt{4x^2+4x+2} - 4\log|(2x+1)+\sqrt{4x^2+4x+2}| + C$

Question 76: Palette 76/107

DPT MCQ

Concept

+

Integral of type $\int \frac{px+q}{x^2+bx+c} \, dx$ with negative discriminant: write $px+q = A\frac{d}{dx}(x^2+bx+c) + B$; first part gives $A\log|x^2+bx+c|$, second part uses completing the square and $\int \frac{du}{u^2+k^2} = \frac{1}{k}\tan^{-1}\left(\frac{u}{k}\right) + C$

Evaluate $\int \frac{x \, dx}{x^2+4x+5}$

A

$\frac{1}{2}\log(x^2+4x+5) + 2\tan^{-1}(x) + C$

B

$\frac{1}{2}\log(x^2+4x+5) - \tan^{-1}(x+2) + C$

C

$\frac{1}{2}\log(x^2+4x+5) + \tan^{-1}(x+2) + C$

D

$\frac{1}{2}\log(x^2+4x+5) - 2\tan^{-1}(x+2) + C$

Question 77: Palette 77/107

DPT MCQ

Concept

+

Integrals of the type $\int \frac{dx}{a\sin^2 x + b\sin x \cos x + c\cos^2 x}$ (or $\int \frac{dx}{a + b\sin^2 x}$, $\int \frac{dx}{a + b\cos^2 x}$) are solved by dividing numerator and denominator by $\cos^2 x$, then substituting $\tan x = t$ so that $\sec^2 x \, dx = dt$, reducing to $\int \frac{dt}{At^2 + Bt + C}$; formula used: $\int \frac{dt}{(t-p)^2 + a^2} = \frac{1}{a}\tan^{-1}\left(\frac{t-p}{a}\right) + C$

Evaluate $\int \frac{dx}{3\sin^2 x + 4\cos^2 x - 6\sin x \cos x}$

A

$\frac{1}{\sqrt{3}}\tan^{-1}\left(\sqrt{3}(\tan x + 1)\right) + C$

B

$\sqrt{3}\tan^{-1}\left(\sqrt{3}(\tan x - 1)\right) + C$

C

$\frac{1}{\sqrt{3}}\tan^{-1}\left(\sqrt{3}(\tan x - 1)\right) + C$

D

$\frac{1}{3}\tan^{-1}\left(\sqrt{3}(\tan x - 1)\right) + C$

Question 78: Palette 78/107

DPT MCQ

Concept

+

Integration of a rational function of $\sin x$ and $\cos x$ by converting to $\tan x$; divide numerator and denominator by $\cos^2 x$, then substitute $t = \tan x$; formulas used: $\sec^2 x = 1 + \tan^2 x$ and $\int \frac{dt}{1 + (kt)^2} = \frac{1}{k} \tan^{-1}(kt) + c$

Evaluate $\int \frac{dx}{1 + 3 \sin^2 x}$

A

$\frac{1}{3} \tan^{-1}(3 \tan x) + c$

B

$\frac{1}{2} \tan^{-1}(2 \tan x) + c$

C

$\tan^{-1}(\tan x) + c$

D

None of these

Question 79: Palette 79/107

DPT MCQ

Concept

+

Substitution $t = \tan x$ so that $dx = \frac{dt}{1 + t^2}$ and $\sin^2 x = \frac{t^2}{1 + t^2}$; formula used: $\int \frac{dt}{a^2 + t^2} = \frac{1}{a} \tan^{-1}\left(\frac{t}{a}\right) + C$

Evaluate $\int \frac{dx}{2 - \sin^2 x}$

A

$\frac{1}{2} \tan^{-1}\left(\frac{\tan x}{2}\right) + C$

B

$\tan^{-1}\left(\frac{\tan x}{\sqrt{2}}\right) + C$

C

$\frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{\tan x}{\sqrt{2}}\right) + C$

D

None of these

Question 80: Palette 80/107

DPT MCQ

Concept

+

Divide numerator and denominator by $\cos^2 x$ to convert to $\tan x$, use $\sec^2 x = 1 + \tan^2 x$, substitute $t = \tan x$; formula used: $\int \frac{dt}{t^2 + a^2} = \frac{1}{a} \tan^{-1}\left(\frac{t}{a}\right) + C$ with $a = 2$

Evaluate $\int \frac{dx}{1 + 3 \cos^2 x}$

A

$\frac{1}{3} \tan^{-1}\left(\frac{\tan x}{3}\right) + C$

B

$\frac{1}{2} \tan^{-1}\left(\frac{\tan x}{2}\right) + C$

C

$2 \tan^{-1}\left(\frac{\tan x}{2}\right) + C$

D

None of these

Question 81: Palette 81/107

DPT MCQ

Concept

+

Divide numerator and denominator by $\cos^2 x$ to convert to $\tan x$, substitute $t = \tan x$; formula used: $\int \frac{dt}{t^2 + a^2} = \frac{1}{a} \tan^{-1}\left(\frac{t}{a}\right) + C$ with $a = \frac{2}{\sqrt{3}}$

Evaluate $\int \frac{dx}{3 \sin^2 x + 4 \cos^2 x}$

A

$\frac{1}{2\sqrt{3}} \tan^{-1}\left(\frac{\sqrt{3} \tan x}{2}\right) + C$

B

$\frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{\sqrt{3} \tan x}{2}\right) + C$

C

$\frac{1}{2\sqrt{3}} \tan^{-1}\left(\frac{2 \tan x}{\sqrt{3}}\right) + C$

D

None of these

Question 82: Palette 82/107

DPT MCQ

Concept

+

Divide numerator and denominator by $\cos^2 x$ to convert to $\tan x$, substitute $t = \tan x$; formula used: $\int \frac{dt}{(at + b)^2} = -\frac{1}{a(at + b)} + C$ with $a = 2$, $b = 3$

Evaluate $\int \frac{dx}{(2 \sin x + 3 \cos x)^2}$

A

$\frac{1}{2(2 \tan x + 3)} + C$

B

$-\frac{1}{2(2 \tan x + 3)} + C$

C

$-\frac{1}{2 \tan x + 3} + C$

D

None of these

Question 83: Palette 83/107

DPT MCQ

Concept

+

Integration of rational functions of $\sin x$ and $\cos x$ by dividing numerator and denominator by $\cos^2 x$, substituting $t = \tan x$, and completing the square in the denominator; formula used: $\int \frac{dt}{(t - \alpha)^2 + a^2} = \frac{1}{a} \tan^{-1}\left(\frac{t - \alpha}{a}\right) + C$

Evaluate $\int \frac{dx}{3 \sin^2 x + 4 \sin x \cos x + 5 \cos^2 x}$

A

$\frac{1}{\sqrt{11}} \tan^{-1}\left(\frac{3 \tan x + 2}{\sqrt{11}}\right) + C$

B

$\frac{3}{\sqrt{11}} \tan^{-1}\left(\frac{3 \tan x + 2}{\sqrt{11}}\right) + C$

C

$\frac{1}{\sqrt{11}} \tan^{-1}\left(\frac{\tan x + 2}{\sqrt{11}}\right) + C$

D

None of these

Question 84: Palette 84/107

DPT MCQ

Concept

+

Reduce to $\tan x$ by dividing numerator and denominator by $\cos^4 x$ (i.e. multiplying by $\sec^4 x$), substitute $t = \tan x$, then divide by $t^2$ and substitute $v = t - \frac{1}{t}$; formula used: $\int \frac{dv}{v^2 + a^2} = \frac{1}{a} \tan^{-1}\left(\frac{v}{a}\right) + C$

Evaluate $I = \int \frac{dx}{\sin^4 x + \cos^4 x - \sin^2 x \cos^2 x}$

A

$\frac{1}{3} \tan^{-1}\left(\frac{\tan x - \frac{1}{\tan x}}{3}\right) + C$

B

$\frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{\tan x - \frac{1}{\tan x}}{\sqrt{3}}\right) + C$

C

$\sqrt{3} \tan^{-1}\left(\frac{\tan x - \frac{1}{\tan x}}{\sqrt{3}}\right) + C$

D

None of these

Question 85: Palette 85/107

DPT MCQ

Concept

+

Expand the square, reduce to $\tan x$ by multiplying by $\sec^4 x$, then substitute $2 \tan x = \tan \theta$; formulas used: $\int \cos^2 \theta \, d\theta = \frac{\theta}{2} + \frac{\sin 2\theta}{4} + C$ and $\sin 2\theta = \frac{2 \tan \theta}{1 + \tan^2 \theta}$

Evaluate $I = \int \frac{dx}{(\cos^2 x + 4 \sin^2 x)^2}$

A

$\frac{5}{16} \tan^{-1}(2 \tan x) - \frac{3 \tan x}{8(1 + 4 \tan^2 x)} + C$

B

$\frac{1}{16} \tan^{-1}(2 \tan x) + \frac{3 \tan x}{8(1 + 4 \tan^2 x)} + C$

C

$\frac{5}{16} \tan^{-1}(2 \tan x) + \frac{3 \tan x}{8(1 + 4 \tan^2 x)} + C$

D

None of these

Question 86: Palette 86/107

DPT MCQ

Concept

+

Weierstrass (tangent half-angle) substitution $t = \tan \frac{x}{2}$, with $\sin x = \frac{2t}{1 + t^2}$, $\cos x = \frac{1 - t^2}{1 + t^2}$, $dx = \frac{2 \, dt}{1 + t^2}$; formula used: $\int \frac{dt}{t + 1} = \ln|t + 1| + C$

Evaluate $\int \frac{1}{1 + \sin x + \cos x} \, dx$

A

$\ln \left| \tan \frac{x}{2} - 1 \right| + C$

B

$\ln \left| \tan \frac{x}{2} + 1 \right| + C$

C

$\ln \left| 1 + \cot \frac{x}{2} \right| + C$

D

None of these

Question 87: Palette 87/107

DPT MCQ

Concept

+

Weierstrass substitution $\tan \frac{x}{2} = t$, with $\cos x = \frac{1 - t^2}{1 + t^2}$, $dx = \frac{2 \, dt}{1 + t^2}$; formula used: $\int \frac{dt}{a^2 - t^2} = \frac{1}{2a} \ln \left| \frac{a + t}{a - t} \right| + C$ with $a = 3$

Integrate $I = \int \frac{dx}{4 + 5 \cos x}$

A

$\frac{1}{6} \ln \left| \frac{3 + \tan \frac{x}{2}}{3 - \tan \frac{x}{2}} \right| + C$

B

$\frac{1}{3} \ln \left| \frac{3 - \tan \frac{x}{2}}{3 + \tan \frac{x}{2}} \right| + C$

C

$\frac{1}{3} \ln \left| \frac{3 + \tan \frac{x}{2}}{3 - \tan \frac{x}{2}} \right| + C$

D

None of these

Question 88: Palette 88/107

DPT MCQ

Concept

+

Weierstrass substitution $\tan \frac{x}{2} = t$, with $\sin x = \frac{2t}{1 + t^2}$, $dx = \frac{2 \, dt}{1 + t^2}$; then partial fractions and $\int \frac{dt}{t + a} = \ln|t + a| + C$

Integrate $\int \frac{dx}{5 - 13 \sin x}$

A

$\frac{1}{24} \ln \left| \frac{\tan \frac{x}{2} - 5}{5 \tan \frac{x}{2} - 1} \right| + C$

B

$\frac{1}{12} \ln \left| \frac{\tan \frac{x}{2} - 5}{5 \tan \frac{x}{2} - 1} \right| + C$

C

$\frac{1}{12} \ln \left| \frac{5 \tan \frac{x}{2} - 1}{\tan \frac{x}{2} - 5} \right| + C$

D

None of these

Question 89: Palette 89/107

DPT MCQ

Concept

+

Weierstrass substitution $\tan \frac{x}{2} = t$, with $\sin x = \frac{2t}{1 + t^2}$, $\cos x = \frac{1 - t^2}{1 + t^2}$, $dx = \frac{2 \, dt}{1 + t^2}$; the quadratic in $t$ becomes a perfect square; formula used: $\int \frac{dt}{(t + a)^2} = -\frac{1}{t + a} + C$

Evaluate $\int \frac{dx}{3 \cos x + 4 \sin x + 5}$

A

$-\frac{1}{\tan \frac{x}{2} + 2} + C$

B

$\frac{1}{\tan \frac{x}{2} + 2} + C$

C

$-\frac{1}{2\left(\tan \frac{x}{2} + 2\right)} + C$

D

None of these

Question 90: Palette 90/107

DPT MCQ

Concept

+

Weierstrass substitution $\tan \frac{x}{2} = t$ with $\cos x = \frac{1 - t^2}{1 + t^2}$, $\sin x = \frac{2t}{1 + t^2}$, $dx = \frac{2 \, dt}{1 + t^2}$; then completing the square and $\int \frac{du}{u^2 + a^2} = \frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right) + C$

Integrate $\int \frac{dx}{13 + 3 \cos x + 4 \sin x}$

A

$\frac{1}{5} \tan^{-1}\left(\frac{5 \tan \frac{x}{2} + 2}{6}\right) + C$

B

$\frac{1}{6} \tan^{-1}\left(\frac{5 \tan \frac{x}{2} - 2}{6}\right) + C$

C

$\frac{1}{6} \tan^{-1}\left(\frac{5 \tan \frac{x}{2} + 2}{6}\right) + C$

D

None of these

Question 91: Palette 91/107

DPT MCQ

Concept

+

Integral of the type $\int \frac{a\cos x + b\sin x + c}{p\cos x + q\sin x + r} dx$ can be solved by expressing the numerator as $(N^r) = A(D^r) + B\frac{d}{dx}(D^r) + C$, where $D^r$ is the denominator. Then $I = \int A dx + B\int \frac{\frac{d}{dx}(D^r)}{D^r} dx + C\int \frac{dx}{D^r} = Ax + B\ln|D^r| + C\int \frac{dx}{D^r}$.

Evaluate $\int \frac{6 + 3\sin x + 14\cos x}{3 + 4\sin x + 5\cos x} dx$

A

$2x - \ln|3 + 4\sin x + 5\cos x| + C$

B

$x + 2\ln|3 + 4\sin x + 5\cos x| + C$

C

$2x + \ln|3 + 4\sin x + 5\cos x| + C$

D

$3x + \ln|3 + 4\sin x + 5\cos x| + C$

Question 92: Palette 92/107

DPT MCQ

Concept

+

Integral of the type $\int \frac{a\cos x + b\sin x + c}{p\cos x + q\sin x + r} dx$ is solved by expressing the numerator as $N = l(D) + m\frac{d}{dx}(D) + n$, giving $I = lx + m\ln|D| + n\int \frac{dx}{D}$. When $r = \sqrt{p^2 + q^2}$, write $p\cos x + q\sin x = r\cos(x - \alpha)$ with $\tan \alpha = \frac{q}{p}$, so $D = 2r\cos^2\left(\frac{x - \alpha}{2}\right)$ and $\int \frac{dx}{D} = \frac{1}{r}\tan\left(\frac{x - \alpha}{2}\right)$.

Evaluate $\int \frac{2\cos x + 3\sin x + 4}{3\cos x + 4\sin x + 5} dx$

A

$\frac{18}{25}x + \frac{1}{25}\ln|3\cos x + 4\sin x + 5| + \frac{2}{25}\tan\left[\frac{x - \tan^{-1}(4/3)}{2}\right] + C$

B

$\frac{18}{25}x - \frac{1}{25}\ln|3\cos x + 4\sin x + 5| - \frac{2}{25}\tan\left[\frac{x - \tan^{-1}(4/3)}{2}\right] + C$

C

$\frac{18}{25}x - \frac{1}{25}\ln|3\cos x + 4\sin x + 5| + \frac{2}{25}\tan\left[\frac{x - \tan^{-1}(3/4)}{2}\right] + C$

D

$\frac{18}{25}x - \frac{1}{25}\ln|3\cos x + 4\sin x + 5| + \frac{2}{25}\tan\left[\frac{x - \tan^{-1}(4/3)}{2}\right] + C$

Question 93: Palette 93/107

DPT MCQ

Concept

+

Integral of the type $\int \frac{a\sin x + b\cos x}{p\sin x + q\cos x} dx$ is solved by expressing the numerator as $N = l(D) + m\frac{d}{dx}(D)$, where $D$ is the denominator; then $I = \int l\, dx + m\int \frac{D'}{D} dx = lx + m\ln|D| + C$, with $l = \frac{ap + bq}{p^2 + q^2}$ and $m = \frac{bp - aq}{p^2 + q^2}$.

Evaluate $\int \frac{2\sin x + 3\cos x}{3\sin x + 4\cos x} dx$

A

$\frac{18}{25}x - \frac{1}{25}\ln|3\sin x + 4\cos x| + C$

B

$\frac{18}{25}x + \frac{1}{25}\ln|3\sin x + 4\cos x| + C$

C

$\frac{18}{25}x + \frac{1}{25}\ln|4\sin x + 3\cos x| + C$

D

$\frac{1}{25}x + \frac{18}{25}\ln|3\sin x + 4\cos x| + C$

Question 94: Palette 94/107

DPT MCQ

Concept

+

Integral of the type $\int \frac{a\sin x + b\cos x + c}{p\sin x + q\cos x + r} dx$ is solved by writing numerator $= \lambda(D) + \mu\frac{d}{dx}(D) + \nu$, so $I = \lambda x + \mu\ln|D| + \nu\int \frac{dx}{D}$; the leftover $\int \frac{dx}{D}$ is evaluated by $t = \tan(x/2)$, giving $\int \frac{2\,dt}{(r - q)t^2 + 2pt + (r + q)}$ after completing the square.

Evaluate $\int \frac{3\cos x + 2}{\sin x + 2\cos x + 3} dx$

A

$\frac{6}{5}x - \frac{3}{5}\ln|\sin x + 2\cos x + 3| - \frac{8}{5}\tan^{-1}\left[\frac{1 + \tan(x/2)}{2}\right] + C$

B

$\frac{6}{5}x + \frac{3}{5}\ln|\sin x + 2\cos x + 3| + \frac{8}{5}\tan^{-1}\left[\frac{1 + \tan(x/2)}{2}\right] + C$

C

$\frac{6}{5}x + \frac{3}{5}\ln|\sin x + 2\cos x + 3| - \frac{8}{5}\tan^{-1}\left[\frac{1 + \tan(x/2)}{2}\right] + C$

D

$\frac{3}{5}x + \frac{6}{5}\ln|\sin x + 2\cos x + 3| - \frac{8}{5}\tan^{-1}\left[\frac{1 + \tan(x/2)}{2}\right] + C$

Question 95: Palette 95/107

DPT MCQ

Concept

+

Integration of a rational exponential function by expressing the numerator in terms of the denominator and its derivative.
Formula: assume $P(x) = A Q(x) + B Q'(x)$, then $\int \frac{P(x)}{Q(x)} \, dx = A \int dx + B \int \frac{Q'(x)}{Q(x)} \, dx = Ax + B \ln|Q(x)| + C$, using $\int \frac{Q'(x)}{Q(x)} \, dx = \ln|Q(x)| + C$.

Evaluate $\int \frac{4e^{x} + 6e^{-x}}{9e^{x} - 4e^{-x}} \, dx$.

A

$\frac{19}{36}x + \frac{35}{36}\ln|9e^{x} - 4e^{-x}| + C$

B

$-\frac{19}{36}x + \frac{35}{36}\ln|9e^{x} - 4e^{-x}| + C$

C

$-\frac{19}{36}x - \frac{35}{36}\ln|9e^{x} - 4e^{-x}| + C$

D

$\frac{35}{36}x - \frac{19}{36}\ln|9e^{x} - 4e^{-x}| + C$

Question 96: Palette 96/107

DPT MCQ

Concept

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For integrals of the type $\int \frac{a e^{x} + b e^{-x}}{p e^{x} + q e^{-x}} \, dx$, express the numerator in terms of the denominator and its derivative.
Formula: let $N^{r} = A \frac{d}{dx}(D^{r}) + B (D^{r})$, then $\int \frac{N^{r}}{D^{r}} \, dx = A \ln|D^{r}| + Bx + C$, using $\int \frac{D'}{D} \, dx = \ln|D| + C$.

Evaluate $\int \frac{3e^{x} + 5e^{-x}}{4e^{x} - 5e^{-x}} \, dx$.

A

$\frac{7}{8}\ln|4e^{x} - 5e^{-x}| + \frac{1}{8}x + C$

B

$\frac{1}{8}\ln|4e^{x} - 5e^{-x}| - \frac{7}{8}x + C$

C

$\frac{7}{8}\ln|4e^{x} - 5e^{-x}| - \frac{1}{8}x + C$

D

$-\frac{7}{8}\ln|4e^{x} - 5e^{-x}| - \frac{1}{8}x + C$

Question 97: Palette 97/107

DPT MCQ

Concept

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Product of three tangents where one angle is the sum of the other two: $3x = 2x + x$.
Formula: $\tan ax \, \tan bx \, \tan cx = \tan ax - \tan bx - \tan cx$ when $a = b + c$, and $\int \tan kx \, dx = \frac{1}{k}\ln|\sec kx| + C$.

Evaluate $\int \tan 3x \, \tan 2x \, \tan x \, dx$.

A

$\frac{1}{3}\ln|\sec 3x| - \frac{1}{2}\ln|\sec 2x| - \ln|\sec x| + C$

B

$\frac{1}{3}\ln|\sec 3x| + \frac{1}{2}\ln|\sec 2x| + \ln|\sec x| + C$

C

$\ln|\sec 3x| - \ln|\sec 2x| - \ln|\sec x| + C$

D

$\frac{1}{3}\ln|\sec 3x| - \frac{1}{2}\ln|\sec 2x| + \ln|\sec x| + C$

Question 98: Palette 98/107

DPT MCQ

Concept

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Express the numerator's angle in terms of the denominator's angle: $x = (x - \alpha) + \alpha$, then expand using $\sin(A + B) = \sin A \cos B + \cos A \sin B$.
Formulas: $\int \cot u \, du = \ln|\sin u| + C$ and $\int dx = x$.

Evaluate $\int \frac{\sin x}{\sin(x - \alpha)} \, dx$.

A

$x \cos \alpha - \sin \alpha \ln|\sin(x - \alpha)| + C$

B

$x \cos \alpha + \sin \alpha \ln|\sin(x - \alpha)| + C$

C

$x \sin \alpha + \cos \alpha \ln|\sin(x - \alpha)| + C$

D

$x \cos \alpha + \sin \alpha \ln|\cos(x - \alpha)| + C$

Question 99: Palette 99/107

DPT MCQ

Concept

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Express the numerator angle as the sum of the denominator angles: $2x = (x - \frac{\pi}{4}) + (x + \frac{\pi}{4})$, then expand $\sin(A + B) = \sin A \cos B + \cos A \sin B$ and split into cotangents.
Formulas: $\int \cot u \, du = \ln|\sin u| + C$.

Evaluate $\int \frac{\sin 2x}{\sin(x + \frac{\pi}{4}) \sin(x - \frac{\pi}{4})} \, dx$.

A

$\ln|\sin(x + \frac{\pi}{4})| - \ln|\sin(x - \frac{\pi}{4})| + C$

B

$\ln|\cos(x + \frac{\pi}{4})| + \ln|\cos(x - \frac{\pi}{4})| + C$

C

$\ln|\sin(x + \frac{\pi}{4})| + \ln|\sin(x - \frac{\pi}{4})| + C$

D

$\frac{1}{2}\ln|\sin(x + \frac{\pi}{4})| + \frac{1}{2}\ln|\sin(x - \frac{\pi}{4})| + C$

Question 100: Palette 100/107

DPT MCQ

Concept

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Multiply and divide by the constant $\sin(a - b)$ and write $\sin(a - b) = \sin\{(x - b) - (x - a)\}$, then expand and split into cotangents.
Formulas: $\int \cot u \, du = \ln|\sin u| + C$.

Evaluate $\int \frac{dx}{\sin(x - a) \sin(x - b)}$.

A

$\frac{1}{\sin(a - b)}[\ln|\sin(x - a)| + \ln|\sin(x - b)|] + C$

B

$\frac{1}{\sin(a - b)}[\ln|\sin(x - a)| - \ln|\sin(x - b)|] + C$

C

$\frac{1}{\sin(a - b)}[\ln|\sin(x - b)| - \ln|\sin(x - a)|] + C$

D

$\frac{1}{\cos(a - b)}[\ln|\sin(x - a)| - \ln|\sin(x - b)|] + C$

Question 101: Palette 101/107

DPT MCQ

Concept

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Multiply and divide by the constant $\cos(a - b) = \cos\{(x - a) - (x - b)\} = \cos(x - a)\cos(x - b) + \sin(x - a)\sin(x - b)$, which splits the integrand into $\cot(x - a) + \tan(x - b)$.
Formulas: $\int \cot u \, du = \ln|\sin u| + C$, $\int \tan u \, du = \ln|\sec u| + C = -\ln|\cos u| + C$.

Evaluate $\int \frac{dx}{\sin(x - a) \cos(x - b)}$.

A

$\frac{1}{\cos(a - b)}[\ln|\sin(x - a)| + \ln|\cos(x - b)|] + C$

B

$\frac{1}{\cos(a - b)}[\ln|\sin(x - a)| - \ln|\cos(x - b)|] + C$

C

$\frac{1}{\sin(a - b)}[\ln|\sin(x - a)| - \ln|\cos(x - b)|] + C$

D

$\frac{1}{\cos(a - b)}[\ln|\cos(x - b)| - \ln|\sin(x - a)|] + C$

Question 102: Palette 102/107

DPT MCQ

Concept

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Integration by Parts formula: $\int u v dx = u \int v dx - \int \left( \frac{du}{dx} \int v dx \right) dx$. Using the ILATE rule, algebraic function $x$ is chosen as first function $u$, and exponential function $e^x$ as second function $v$.

Evaluate $\int x e^x dx$

A

$x e^x + e^x + C$

B

$x e^x - e^x + C$

C

$x^2 e^x - e^x + C$

D

$e^x - x e^x + C$

Question 103: Palette 103/107

DPT MCQ

Concept

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Integration by Parts formula: $\int u v dx = u \int v dx - \int \left( \frac{du}{dx} \int v dx \right) dx$. Here, set $u = \tan^{-1} x$ as the inverse trigonometric function and $v = 1$ as the algebraic function.

Evaluate $\int \tan^{-1} x dx$

A

$x \tan^{-1} x + \frac{1}{2} \ln(1 + x^2) + C$

B

$x \tan^{-1} x - \frac{1}{2} \ln(1 + x^2) + C$

C

$x \tan^{-1} x - \ln(1 + x^2) + C$

D

$\frac{x^2}{2} \tan^{-1} x - \frac{x}{2} + C$

Question 104: Palette 104/107

DPT MCQ

Concept

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Integration by Parts formula: $\int u v dx = u \int v dx - \int \left( \frac{du}{dx} \int v dx \right) dx$. Using the ILATE rule, choose $u = \ln(1 + x)$ as the logarithmic function and $v = x$ as the algebraic function.

Evaluate $\int x \ln(1 + x) dx$

A

$\frac{x^2}{2} \ln(1 + x) - \frac{x^2}{4} + \frac{x}{2} - \frac{1}{2} \ln\vert{}1 + x\vert{} + C$

B

$\frac{x^2 - 1}{2} \ln(1 + x) - \frac{x^2}{4} + \frac{x}{2} + C$

C

$\frac{x^2}{2} \ln(1 + x) - \frac{1}{2} \left[ \frac{x^2}{2} - x + \ln\vert{}1 + x\vert{} \right] + C$

D

$\frac{x^2}{2} \ln(1 + x) + \frac{x^2}{4} - \frac{x}{2} + \ln\vert{}1 + x\vert{} + C$

Question 105: Palette 105/107

DPT MCQ

Concept

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Integration by Parts formula: $\int u v dx = u \int v dx - \int \left( \frac{du}{dx} \int v dx \right) dx$. Here, set $u = \ln(x + \sqrt{x^2 + a^2})$ as the logarithmic function and $v = 1$ as unity.

Evaluate $\int \ln(x + \sqrt{x^2 + a^2}) dx$

A

$x \ln(x + \sqrt{x^2 + a^2}) + \sqrt{x^2 + a^2} + C$

B

$x \ln(x + \sqrt{x^2 + a^2}) - \sqrt{x^2 + a^2} + C$

C

$\sqrt{x^2 + a^2} \ln(x + \sqrt{x^2 + a^2}) - x + C$

D

$x \ln(x + \sqrt{x^2 + a^2}) - \frac{1}{2}\sqrt{x^2 + a^2} + C$

Question 106: Palette 106/107

DPT MCQ

Concept

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Integration by Parts formula: $\int u \, dv = u v - \int v \, du$. Using ILATE rule, algebraic function $u = x^2$ is chosen as the first function and exponential function $v' = e^{-x}$ as the second function.

Find $\int x^2 e^{-x} \, dx$

A

$-e^{-x} (x^2 + 2x + 2) + C$

B

$e^{-x} (x^2 - 2x + 2) + C$

C

$-e^{-x} (x^2 - 2x + 2) + C$

D

$-e^{-x} (x^2 + 2x - 2) + C$

Question 107: Palette 107/107

DPT MCQ

Concept

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Integration by Parts formula: $\int u \, dv = u v - \int v \, du$. Using the ILATE rule, choose $u = x^3$ as the algebraic function and $dv = e^{2x} dx$ as the exponential function, applying repeated integration by parts to reduce the polynomial degree.

Evaluate $I = \int x^3 e^{2x} \, dx$

A

$\frac{1}{8} e^{2x} (4x^3 - 6x^2 + 6x - 3) + C$

B

$\frac{1}{8} e^{2x} (4x^3 + 6x^2 + 6x + 3) + C$

C

$\frac{1}{4} e^{2x} (2x^3 - 3x^2 + 3x - 3) + C$

D

$\frac{1}{8} e^{2x} (4x^3 - 6x^2 - 6x + 3) + C$