iCON Education - Integral Calculus

Q1 DPT Indefinite Integration MCQ

iCON Education HYD, 79930 92826, 73309 72826 14 Sep 2026

Concept:

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.

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

Q2 DPT Indefinite Integration MCQ

iCON Education HYD, 79930 92826, 73309 72826 14 Sep 2026

Concept:

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

A.

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

B.

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

C.

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

D.

None of these

Q3 DPT Indefinite Integration MCQ

iCON Education HYD, 79930 92826, 73309 72826 14 Sep 2026

Concept:

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

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.

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

Correct Answer: A

Explanation:

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

Q4 DPT Indefinite Integration MCQ

iCON Education HYD, 79930 92826, 73309 72826 14 Sep 2026

Concept:

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

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

Q5 DPT Indefinite Integration MCQ

iCON Education HYD, 79930 92826, 73309 72826 14 Sep 2026

Concept:

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

A.

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

B.

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

C.

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

D.

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

Correct Answer: A

Explanation:

$\int \frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha} \, dx$
$= \int \frac{(2\cos^2 x - 1) - (2\cos^2 \alpha - 1)}{\cos x - \cos \alpha} \, dx$
$= \int \frac{2[\cos^2 x - \cos^2 \alpha]}{\cos x - \cos \alpha} \, dx = 2 \int (\cos x + \cos \alpha) \, dx$
$= 2\left[\int \cos x \, dx + \int \cos \alpha \, dx\right] = 2\left[\sin x + \cos \alpha \int dx\right]$
$= 2[\sin x + x \cos \alpha] + c$

Q6 DPT Indefinite Integration MCQ

iCON Education HYD, 79930 92826, 73309 72826 14 Sep 2026

Concept:

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

A.

$\tan x + \cot x + c$

B.

$\sec x \tan x + c$

C.

$\csc x \cot x + c$

D.

None of these

Correct Answer: D

Explanation:

$\int (\tan x - \cot x)^2 \, dx = \int [\tan^2 x + \cot^2 x - 2] \, dx$
$= \int \tan^2 x \, dx + \int \cot^2 x \, dx - \int 2 \, dx$
$= \int (\sec^2 x - 1) \, dx + \int (\csc^2 x - 1) \, dx - 2x + c$
$= \int \sec^2 x \, dx + \int \csc^2 x \, dx - 4x + c$
$= \tan x - \cot x - 4x + c$

Q7 DPT Indefinite Integration MCQ

iCON Education HYD, 79930 92826, 73309 72826 14 Sep 2026

Concept:

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

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.

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

Correct Answer: A

Explanation:

$\int (\sec x + \tan x)^2 \, dx$
$= \int (\sec^2 x + \tan^2 x + 2\sec x \tan x) \, dx$
$\because \sec^2 x - 1 = \tan^2 x$
$\therefore I = 2 \int \sec^2 x \, dx - 1 \int dx + \int 2\sec x \tan x \, dx$
$= 2\tan x - x + 2\sec x + c$

Q8 DPT Indefinite Integration MCQ

iCON Education HYD, 79930 92826, 73309 72826 14 Sep 2026

Concept:

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

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

Correct Answer: A

Explanation:

$\int \frac{\csc\theta - \cot\theta}{\csc\theta + \cot\theta} \, d\theta$
$= \int \frac{(\csc\theta - \cot\theta)^2}{\csc^2\theta - \cot^2\theta} \, d\theta$
$= \int (\csc\theta - \cot\theta)^2 \, d\theta$
$= \int (\csc^2\theta + \cot^2\theta - 2\csc\theta\cot\theta) \, d\theta$
$= \int (\csc^2\theta + \csc^2\theta - 1 - 2\csc\theta\cot\theta) \, d\theta$
$= \int (2\csc^2\theta - 2\csc\theta\cot\theta - 1) \, d\theta$
$= -2\cot\theta + 2\csc\theta - \theta + c$
$= 2\csc\theta - 2\cot\theta - \theta + c$

Q9 DPT Indefinite Integration MCQ

iCON Education HYD, 79930 92826, 73309 72826 14 Sep 2026

Concept:

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

A.

$x + \cos x + c$

B.

$1 + \sin x + c$

C.

$\sec x - \tan x + c$

D.

$\sec x + \tan x + c$

Correct Answer: D

Explanation:

$\int \frac{dx}{1 - \sin x} = \int \frac{1 + \sin x}{(1 - \sin x)(1 + \sin x)} \, dx$
$= \int \frac{1 + \sin x}{\cos^2 x} \, dx = \int \left( \sec^2 x + \frac{\sin x}{\cos^2 x} \right) \, dx$
$= \int \sec^2 x \, dx + \int \frac{\sin x}{\cos x \cdot \cos x} \, dx$
$= \int \sec^2 x \, dx + \int \tan x \sec x \, dx = \tan x + \sec x + c$

Q10 DPT Indefinite Integration MCQ

iCON Education HYD, 79930 92826, 73309 72826 14 Sep 2026

Concept:

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

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.

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

Q11 DPT Indefinite Integration MCQ

iCON Education HYD, 79930 92826, 73309 72826 14 Sep 2026

Concept:

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

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.

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

Correct Answer: D

Explanation:

$\because 1 + \sin x = \left(\cos\frac{x}{2} + \sin\frac{x}{2}\right)^2$
$\therefore \int \sqrt{1 + \sin x} \, dx = \int \sqrt{\left(\cos\frac{x}{2} + \sin\frac{x}{2}\right)^2} \, dx$
$= \int \left(\cos\frac{x}{2} + \sin\frac{x}{2}\right) \, dx = \frac{\sin\frac{x}{2}}{\frac{1}{2}} - \frac{\cos\frac{x}{2}}{\frac{1}{2}} + c$
$= 2\left(\sin\frac{x}{2} - \cos\frac{x}{2}\right) + c = -2\sqrt{1 - \sin x} + c$

Q12 DPT Indefinite Integration MCQ

iCON Education HYD, 79930 92826, 73309 72826 14 Sep 2026

Concept:

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

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.

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

Q13 DPT Indefinite Integration MCQ

iCON Education HYD, 79930 92826, 73309 72826 14 Sep 2026

Concept:

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

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.

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

Q14 DPT Indefinite Integration MCQ

iCON Education HYD, 79930 92826, 73309 72826 14 Sep 2026

Concept:

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

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

Correct Answer: A

Explanation:

$\cos^4 x = (\cos^2 x)^2 = \left(\frac{1 + \cos 2x}{2}\right)^2 = \frac{1 + 2\cos 2x + \cos^2 2x}{4}$
$= \frac{1 + 2\cos 2x + \frac{1 + \cos 4x}{2}}{4} = \frac{3 + 4\cos 2x + \cos 4x}{8}$
$\int \cos^4 x \, dx = \int \frac{3 + 4\cos 2x + \cos 4x}{8} \, dx = \frac{3x}{8} + \frac{\sin 2x}{4} + \frac{\sin 4x}{32} + c$

Q15 DPT Indefinite Integration MCQ

iCON Education HYD, 79930 92826, 73309 72826 14 Sep 2026

Concept:

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

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.

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

Q16 DPT Indefinite Integration MCQ

iCON Education HYD, 79930 92826, 73309 72826 14 Sep 2026

Concept:

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

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

Correct Answer: A

Explanation:

$\sin^3 x \cos^3 x = (\sin x \cos x)^3 = \left(\frac{\sin 2x}{2}\right)^3 = \frac{\sin^3 2x}{8}$
$\because \sin^3 2x = \frac{3\sin 2x - \sin 6x}{4}$
$\therefore \int \sin^3 x \cos^3 x \, dx = \frac{1}{32} \int (3\sin 2x - \sin 6x) \, dx$
$= \frac{1}{32} \left[ -\frac{3\cos 2x}{2} + \frac{\cos 6x}{6} \right] + c$

Q17 DPT Indefinite Integration MCQ

iCON Education HYD, 79930 92826, 73309 72826 14 Sep 2026

Concept:

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

A.

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

B.

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

C.

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

D.

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

Correct Answer: D

Explanation:

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

Q18 DPT Indefinite Integration MCQ

iCON Education HYD, 79930 92826, 73309 72826 14 Sep 2026

Concept:

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

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

Q19 DPT Indefinite Integration MCQ

iCON Education HYD, 79930 92826, 73309 72826 14 Sep 2026

Concept:

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

Correct Answer: B

Explanation:

$\int (1 + 2x + 3x^2 + 4x^3 + \dots) \, dx$
$= x + x^2 + x^3 + x^4 + \dots \infty = \frac{x}{1 - x} \quad \left(\because S_\infty = \frac{a}{1 - r}\right)$
$= \frac{x - 1 + 1}{1 - x} + c = \left( \frac{x - 1}{1 - x} \right) + \frac{1}{1 - x} + c$
$= -1 + \frac{1}{1 - x} + c = \frac{1}{1 - x} + c' \quad (c' = c - 1)$

Q20 DPT Indefinite Integration MCQ

iCON Education HYD, 79930 92826, 73309 72826 14 Sep 2026

Concept:

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

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

Correct Answer: C

Explanation:

$\int \sin 2x \cos^2 x \, dx = \int \sin 2x \frac{(1 + \cos 2x)}{2} \, dx$
$= \frac{1}{2} \int [\sin 2x + \sin 2x \cos 2x] \, dx$
$= \frac{1}{2} \left[ \int \sin 2x \, dx + \int \sin 2x \cos 2x \, dx \right]$
$= \frac{1}{2} \left[ -\frac{\cos 2x}{2} + \frac{1}{2} \int \sin 4x \, dx \right]$
$= -\frac{\cos 2x}{4} - \frac{\cos 4x}{16} + c$

Q21 DPT Indefinite Integration MCQ

iCON Education HYD, 79930 92826, 73309 72826 14 Sep 2026

Concept:

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

A.

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

B.

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

C.

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

D.

None of these

Q22 DPT Indefinite Integration MCQ

iCON Education HYD, 79930 92826, 73309 72826 14 Sep 2026

Concept:

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

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

Correct Answer: C

Explanation:

$\int \frac{2 + 3x^2}{x^2(1 + x^2)} \, dx = \int \frac{2 + 2x^2 + x^2}{x^2(1 + x^2)} \, dx$
$= \int \frac{2(1 + x^2) + x^2}{x^2(1 + x^2)} \, dx = \int \left[ \frac{2}{x^2} + \frac{1}{1 + x^2} \right] \, dx$
$= \int 2x^{-2} \, dx + \int \frac{dx}{1 + x^2}$
$= \frac{2x^{-2+1}}{-2+1} + \tan^{-1} x + c = -\frac{2}{x} + \tan^{-1} x + c$

Q23 DPT Indefinite Integration MCQ

iCON Education HYD, 79930 92826, 73309 72826 14 Sep 2026

Concept:

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

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

Correct Answer: B

Explanation:

$\int \frac{(x^4 + x^2) + 1}{2(1 + x^2)} \, dx = \int \frac{x^2(x^2 + 1) + 1}{2(1 + x^2)} \, dx$
$= \int \left[ \frac{x^2(x^2 + 1)}{2(1 + x^2)} + \frac{1}{2(1 + x^2)} \right] \, dx$
$= \frac{1}{2} \int x^2 \, dx + \frac{1}{2} \int \frac{dx}{1 + x^2}$
$= \frac{1}{2} \cdot \frac{x^3}{3} + \frac{1}{2}\tan^{-1} x + c = \frac{x^3}{6} + \frac{1}{2}\tan^{-1} x + c$

Q24 DPT Indefinite Integration MCQ

iCON Education HYD, 79930 92826, 73309 72826 14 Sep 2026

Concept:

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

A.

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

B.

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

C.

$x^2 - x + c$

D.

None of these

Correct Answer: A

Explanation:

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

Q25 DPT Indefinite Integration MCQ

iCON Education HYD, 79930 92826, 73309 72826 14 Sep 2026

Concept:

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

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

Q26 DPT Indefinite Integration MCQ

iCON Education HYD, 79930 92826, 73309 72826 14 Sep 2026

Concept:

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

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

Correct Answer: C

Explanation:

$\int \frac{\sin 2x - \sin 2k}{\sin x - \sin k + \cos x - \cos k} \, dx$
$= \int \frac{(1 + \sin 2x) - (1 + \sin 2k)}{(\sin x - \sin k) + (\cos x - \cos k)} \, dx$
$= \int \frac{(\sin x + \cos x)^2 - (\sin k + \cos k)^2}{\sin x - \sin k + \cos x - \cos k} \, dx$
$= \int \frac{(\sin x + \cos x + \sin k + \cos k)(\sin x + \cos x - \sin k - \cos k)}{\sin x - \sin k + \cos x - \cos k} \, dx$
$= \int \sin x \, dx + \int \cos x \, dx + \int \sin k \, dx + \int \cos k \, dx$
$= -\cos x + \sin x + x\sin k + x\cos k + c$

Q27 DPT Indefinite Integration MCQ

iCON Education HYD, 79930 92826, 73309 72826 14 Sep 2026

Concept:

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

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

Correct Answer: B

Explanation:

$\int \frac{x^2 + 3}{x^6(x^2 + 1)} \, dx = \int \frac{(x^2 + 1) + 2}{x^6(x^2 + 1)} \, dx = \int \left[ \frac{1}{x^6} + \frac{2}{x^6(x^2 + 1)} \right] \, dx$
$= \int \frac{1}{x^6} \, dx + 2 \int \left[ \frac{1}{x^6} - \frac{1}{x^4} + \frac{1}{x^2} - \frac{1}{x^2 + 1} \right] \, dx$
$= 3 \int x^{-6} \, dx - 2 \int x^{-4} \, dx + 2 \int x^{-2} \, dx - 2 \int \frac{dx}{x^2 + 1}$
$= 3\left(\frac{x^{-5}}{-5}\right) - 2\left(\frac{x^{-3}}{-3}\right) + 2\left(\frac{x^{-1}}{-1}\right) - 2\tan^{-1} x + c$
$= -\frac{3}{5x^5} + \frac{2}{3x^3} - \frac{2}{x} - 2\tan^{-1} x + c$

Q28 DPT Integration by Substitution MCQ

iCON Education HYD, 79930 92826, 73309 72826 17 Sep 2026

Concept:

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.

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

Q29 DPT Integration by Substitution MCQ

iCON Education HYD, 79930 92826, 73309 72826 17 Sep 2026

Concept:

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.

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

Q30 DPT Integration by Substitution MCQ

iCON Education HYD, 79930 92826, 73309 72826 17 Sep 2026

Concept:

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.

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

Q31 DPT Integration by Substitution MCQ

iCON Education HYD, 79930 92826, 73309 72826 17 Sep 2026

Concept:

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.

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

Q32 DPT Integration by Substitution MCQ

iCON Education HYD, 79930 92826, 73309 72826 17 Sep 2026

Concept:

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.

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

Correct Answer: A

Explanation:

$\int \frac{a^x}{\sqrt{1 - a^{2x}}} dx$
Let $a^x = t \Rightarrow a^x \ln a \cdot dx = dt \Rightarrow a^x dx = \frac{dt}{\ln a}$
$I = \int \frac{dt}{\ln a \sqrt{1 - t^2}} = \frac{1}{\ln a} \int \frac{dt}{\sqrt{1 - t^2}} = \frac{1}{\ln a} \sin^{-1}(a^x) + c$

Q33 DPT Integration by Substitution MCQ

iCON Education HYD, 79930 92826, 73309 72826 17 Sep 2026

Concept:

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

Q34 DPT Integration by Substitution MCQ

iCON Education HYD, 79930 92826, 73309 72826 17 Sep 2026

Concept:

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.

$6\sqrt{3}$

Correct Answer: C

Explanation:

$I(x) = \int \frac{6(\csc^2 x dx)}{(1 - \cot x)^2}$
let, $1 - \cot x = t \Rightarrow \csc^2 x dx = dt$
$I = 6 \int \frac{dt}{t^2} = 6 \int t^{-2} dt = \frac{6t^{-2+1}}{-2+1} = -\frac{6}{t} + c$
$\Rightarrow I(x) = \frac{-6}{1 - \cot x} + c$
Given, at $x = 0 \Rightarrow I(0) = 3$
$3 = \frac{-6}{1 - \cot 0} + c \Rightarrow c = 3$
$I(x) = \frac{-6}{1 - \cot x} + 3$
$I\left(\frac{\pi}{12}\right) = \frac{-6}{1 - \cot\left(\frac{\pi}{12}\right)} + 3 = \frac{-6}{1 - (2 + \sqrt{3})} + 3 = \frac{-6}{-1 - \sqrt{3}} + 3 = 3\sqrt{3} - 3 + 3 = 3\sqrt{3}$

Q35 DPT Integration by Substitution MCQ

iCON Education HYD, 79930 92826, 73309 72826 17 Sep 2026

Concept:

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.

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

Correct Answer: B

Explanation:

$I = \int \frac{\sqrt{\tan x}}{\frac{\sin x}{\cos x}(\cos^2 x)} dx = \int \frac{\sqrt{\tan x}}{\tan x} \sec^2 x dx$
$= \int \frac{\sec^2 x dx}{\sqrt{\tan x}}$
let, $\tan x = t^2 \Rightarrow \sec^2 x dx = 2t dt$
$I = \int \frac{2t dt}{t} = 2t + c = 2\sqrt{\tan x} + c$

Q36 DPT Integration by Substitution MCQ

iCON Education HYD, 79930 92826, 73309 72826 17 Sep 2026

Concept:

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

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

Q37 DPT Integration by Substitution MCQ

iCON Education HYD, 79930 92826, 73309 72826 17 Sep 2026

Concept:

$\int \cos^3 x dx$

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

Q38 DPT Integration by Substitution MCQ

iCON Education HYD, 79930 92826, 73309 72826 17 Sep 2026

Concept:

$\int \cos^5 x dx$

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

Q39 DPT Integration by Substitution MCQ

iCON Education HYD, 79930 92826, 73309 72826 17 Sep 2026

Concept:

$\int \cos^4 x dx$

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

Correct Answer: A

Explanation:

$I = \int \cos^3 x \cos x dx = \int (1 - \sin^2 x)^2 \cos x dx$
let, $\sin x = t, I = \int (1 - t^2)^2 dt \rightarrow \text{stuck}$
$I = \int \left(\frac{1 + \cos 2x}{2}\right)^2 dx = \frac{1}{4} \int (1 + 2\cos 2x + \cos^2 2x) dx$
$= \frac{1}{4} \int \left(1 + 2\cos 2x + \frac{1 + \cos 4x}{2}\right) dx = \frac{1}{4} \int \left(\frac{3}{2} + 2\cos 2x + \frac{\cos 4x}{2}\right) dx$
$= \frac{3x}{8} + \frac{\sin 2x}{4} + \frac{\sin 4x}{32} + c$

Q40 DPT Integration by Substitution MCQ

iCON Education HYD, 79930 92826, 73309 72826 17 Sep 2026

Concept:

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

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

Q41 DPT Integration by Substitution MCQ

iCON Education HYD, 79930 92826, 73309 72826 17 Sep 2026

Concept:

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

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

Q42 DPT Integration by Substitution MCQ

iCON Education HYD, 79930 92826, 73309 72826 17 Sep 2026

Concept:

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

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

Q43 DPT Integration by Substitution MCQ

iCON Education HYD, 79930 92826, 73309 72826 17 Sep 2026

Concept:

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

A.

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

B.

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

C.

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

D.

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

Q44 DPT Integration by Substitution MCQ

iCON Education HYD, 79930 92826, 73309 72826 17 Sep 2026

Concept:

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

Correct Answer: A

Explanation:

$I = \frac{1}{\sin(a - b)} \int \frac{\sin(a - b)}{\cos(x - a)\cos(x - b)} dx$
$= \frac{1}{\sin(a - b)} \int \frac{\sin((x - b) - (x - a))}{\cos(x - a)\cos(x - b)} dx$
$= \frac{1}{\sin(a - b)} \int \frac{\sin(x - b)\cos(x - a) - \cos(x - b)\sin(x - a)}{\cos(x - a)\cos(x - b)} dx$
$= \frac{1}{\sin(a - b)} \int (\tan(x - b) - \tan(x - a)) dx$
$= \frac{1}{\sin(a - b)} [\ln\vert{}\sec(x - b)\vert{} - \ln\vert{}\sec(x - a)\vert{}] + c$
$= \frac{1}{\sin(a - b)} \ln\left\vert{} \frac{\sec(x - b)}{\sec(x - a)} \right\vert{} + c = \frac{1}{\sin(a - b)} \ln\left\vert{} \frac{\cos(x - a)}{\cos(x - b)} \right\vert{} + c$

Q45 DPT Integration by Substitution MCQ

iCON Education HYD, 79930 92826, 73309 72826 17 Sep 2026

Concept:

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

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

Q46 DPT Integration by Substitution MCQ

iCON Education HYD, 79930 92826, 73309 72826 17 Sep 2026

Concept:

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

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

Q47 DPT Integration by Substitution MCQ

iCON Education HYD, 79930 92826, 73309 72826 17 Sep 2026

Concept:

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

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

Correct Answer: A

Explanation:

$I = \frac{1}{\sin(a - b)} \int \frac{\sin(a - b)}{\cos(x - a)\cos(x - b)} dx$
$= \frac{1}{\sin(a - b)} \int \frac{\sin((x - b) - (x - a))}{\cos(x - a)\cos(x - b)} dx$
$= \frac{1}{\sin(a - b)} \int \frac{\sin(x - b)\cos(x - a) - \cos(x - b)\sin(x - a)}{\cos(x - a)\cos(x - b)} dx$
$= \frac{1}{\sin(a - b)} \int (\tan(x - b) - \tan(x - a)) dx$
$= \frac{1}{\sin(a - b)} [\ln\vert{}\sec(x - b)\vert{} - \ln\vert{}\sec(x - a)\vert{}] + c$
$= \frac{1}{\sin(a - b)} \ln\left\vert{} \frac{\sec(x - b)}{\sec(x - a)} \right\vert{} + c = \frac{1}{\sin(a - b)} \ln\left\vert{} \frac{\cos(x - a)}{\cos(x - b)} \right\vert{} + c$

Q48 DPT Integration by Tricks MCQ

iCON Education HYD, 79930 92826, 73309 72826 18 Sep 2026

Concept:

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.

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

Q49 DPT Integration by Tricks MCQ

iCON Education HYD, 79930 92826, 73309 72826 18 Sep 2026

Concept:

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.

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

Q50 DPT Integration by Tricks MCQ

iCON Education HYD, 79930 92826, 73309 72826 18 Sep 2026

Concept:

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.

$e^{2x} + C$

Mathematics-II Chapters

Integral Calculus

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