Indefinite Integrals

Given

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

$f(0)=-26+24 \ln 2 \text { and } $

$ f(1)=a+b \ln 3, \quad a, b \in \mathbb{Z} $

Use substitution Let $x=t^6 \Longrightarrow d x=6 t^5 d t$

$ \begin{aligned} & x^{2 / 3}=\left(t^6\right)^{2 / 3}=t^4 \\ & x^{1 / 2}=\left(t^6\right)^{1 / 2}=t^3 \end{aligned} $

Substituting these into the integral :

$ f(x)=\int \frac{6 t^5}{t^4+2 t^3} d t=\int \frac{6 t^5}{t^3(t+2)} d t=\int \frac{6 t^2}{t+2} d t $

manipulate

$ \frac{t^2}{t+2}=\frac{\left(t^2-4\right)+4}{t+2}=\frac{(t-2)(t+2)+4}{t+2}=(t-2)+\frac{4}{t+2} $

Now, integrate term by term:

$ \begin{gathered} f(x)=6 \int\left(t-2+\frac{4}{t+2}\right) d t \\ f(x)=6\left[\frac{t^2}{2}-2 t+4 \ln |t+2|\right]+C \end{gathered} $

$f(x)=3 t^2-12 t+24 \ln |t+2|+C$

Back-substitute $t=x^{1 / 6}$ :

$ f(x)=3 x^{1 / 3}-12 x^{1 / 6}+24 \ln \left|x^{1 / 6}+2\right|+C $

Finding the Constant $C$

We are given the initial condition $f(0)=-26+24 \ln (2)$ :

$ \begin{gathered} f(0)=3(0)-12(0)+24 \ln (0+2)+C \\ -26+24 \ln (2)=24 \ln (2)+C \Longrightarrow \mathbf{C}=-\mathbf{2 6} \end{gathered} $

So, the complete function is:

$ f(x)=3 x^{1 / 3}-12 x^{1 / 6}+24 \ln \left(x^{1 / 6}+2\right)-26 $

Solving for $a$ and $b$

Now we evaluate $f(1)$ :

$ \begin{gathered} f(1)=3(1)^{1 / 3}-12(1)^{1 / 6}+24 \ln \left(1^{1 / 6}+2\right)-26 \\ f(1)=3-12+24 \ln (3)-26 \\ f(1)=-35+24 \ln (3) \end{gathered} $

Comparing this to the given form $f(1)=a+b \ln (3)$ :

$ a=-35 \quad \text { and } \quad b=24 $

Then

$ a+b=-35+24=-\mathbf{1 1}$

The given integral is :

$ I=\int\left(\frac{1-5 \cos ^2 x}{\sin ^5 x \cos ^2 x}\right) d x $

We can rewrite the integrand by splitting the terms:

$ \begin{gathered} I=\int\left(\frac{1}{\sin ^5 x \cos ^2 x}-\frac{5 \cos ^2 x}{\sin ^5 x \cos ^2 x}\right) d x \\ I=\int\left(\frac{\sec ^2 x}{\sin ^5 x}-5 \operatorname{cosec}^5 x\right) d x \end{gathered} $

Observe that the derivative of $f(x)=\frac{\tan x}{\sin ^5 x}$ gives us this exact expression:

$ \begin{aligned} & \frac{d}{d x}\left(\frac{\tan x}{\sin ^5 x}\right)=\frac{\sec ^2 x \cdot \sin ^5 x-\tan x \cdot\left(5 \sin ^4 x \cos x\right)}{\sin ^{10} x} \\ & =\frac{\sec ^2 x \sin ^5 x-5 \sin ^5 x}{\sin ^{10} x}=\frac{\sec ^2 x-5}{\sin ^5 x}=\frac{1-5 \cos ^2 x}{\sin ^5 x \cos ^2 x} \end{aligned} $

Therefore, $f(x)=\frac{\tan x}{\sin ^5 x}=\frac{\sin x / \cos x}{\sin ^5 x}=\frac{1}{\sin ^4 x \cos x}$.

Calculate $f\left(\frac{\pi}{6}\right)-f\left(\frac{\pi}{4}\right)$

We substitute the values into the function $f(x)=\frac{1}{\sin ^4 x \cos x}$ :

• At $x=\frac{\pi}{6}$ :

$ \begin{gathered} \sin \left(\frac{\pi}{6}\right)=\frac{1}{2}, \quad \cos \left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2} \\ f\left(\frac{\pi}{6}\right)=\frac{1}{(1 / 2)^4 \cdot(\sqrt{3} / 2)}=\frac{1}{(1 / 16) \cdot(\sqrt{3} / 2)}=\frac{32}{\sqrt{3}} \end{gathered} $

• At $x=\frac{\pi}{4}$ :

$\sin \left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}, \quad \cos \left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}} $

$\Rightarrow $$ f\left(\frac{\pi}{4}\right)=\frac{1}{(1 / \sqrt{2})^4 \cdot(1 / \sqrt{2})}=\frac{1}{(1 / 4) \cdot(1 / \sqrt{2})}=4 \sqrt{2} $

$\therefore $ $ f\left(\frac{\pi}{6}\right)-f\left(\frac{\pi}{4}\right)=\frac{32}{\sqrt{3}}-4 \sqrt{2}$

Factor out $\frac{4}{\sqrt{3}}$ to match the options :

$ =\frac{4}{\sqrt{3}}(8-\sqrt{2} \cdot \sqrt{3})=\frac{4}{\sqrt{3}}(8-\sqrt{6}) $

$ f(t)=\int\left(\frac{1-\sin \left(\log _e t\right)}{1-\cos \left(\log _e t\right)}\right) d t, t>1 $

Let $\log _e t=x$.

$ \begin{aligned} & t=e^x \Rightarrow d t=e^x d x \\ & f(t)=\int\left(\frac{1-\sin x}{1-\cos x}\right) e^x d x \\ & =\int\left(\frac{1-2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin ^2 \frac{x}{2}}\right) e^x d x\left\{\begin{array}{l} \text { use } \sin 2 \theta=2 \sin \theta \cos \theta \\ 1-\cos 2 \theta=2 \sin ^2 \theta \end{array}\right\} \\ & f(t)=\int\left[\frac{1}{2} \operatorname{cosec}^2 \frac{x}{2}-\cot \frac{x}{2}\right] e^x d x \end{aligned} $

This standard integral of type $\int e^x\left[g(x)+g^{\prime}(x)\right]=e^x \cdot g(x)+C$.

let $g(x)=-\cot \frac{x}{2}$.

$ g^{\prime}(x)=-\left(-\operatorname{cosec}^2 \frac{x}{2}\right) \cdot \frac{1}{2}=\frac{1}{2} \operatorname{cosec}^2 \frac{x}{2} . $

so $f(t)=e^x \cdot\left(-\cot \frac{x}{2}\right)+C$.

substitute value of $x=\log _e t$ back.

$ f(t)=-t \cot \left(\frac{\log _e t}{2}\right)+C $

use $f\left(e^{\pi / 2}\right)=-e^{\pi / 2}$ to find C :

$ \begin{aligned} & -e^{\pi / 2}=-e^{\pi / 2} \cot \left(\frac{\log _e e^{\pi / 2}}{2}\right)+C \\ & -e^{\pi / 2}=-e^{\pi / 2} \cot \left(\frac{\pi}{4}\right)+C \\ & -e^{\pi / 2}=-e^{\pi / 2}+C \Rightarrow C=0 \end{aligned} $

$ f(t)=-t \cot \left(\frac{\log _e t}{2}\right) $

It is given that $f\left(e^{\pi / 4}\right)=\alpha e^{\pi / 4}$.

$ \begin{aligned} & \alpha e^{\pi / 4}=-e^{\pi / 4} \cot \left(\frac{\log _e e^{\pi / 4}}{2}\right) \\ & \alpha=-\cot \left(\frac{\pi}{8}\right) \end{aligned} $

using result $1+\cos 2 \theta=2 \cos ^2 \theta\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(1)$

$ \begin{aligned} & \sin 2 \theta=2 \sin \theta \cos \theta \\ & \frac{1+\cos 2 \theta}{\sin 2 \theta}=\frac{2 \cos ^2 \theta}{2 \sin \theta \cos \theta}=\cot \theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(2)\end{aligned} $

$ \alpha=-\cot \left(\frac{\pi}{8}\right)=-\left(\frac{1+\cos \frac{\pi}{4}}{\sin \frac{\pi}{4}}\right) $

$\alpha=-\left(\frac{1+\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}\right)=-(\sqrt{2}+1)$.

$I(x)=\int \frac{3 d x}{(4 x+6) \sqrt{4 x^2+8 x+3}}=\int \frac{3 d x}{2(2 x+3) \sqrt{(2 x+2)^2-1}}$

let $2 x+3=\frac{1}{t} \Longrightarrow 2 d x=-\frac{1}{t^2} d t \Longrightarrow d x=-\frac{1}{2 t^2} d t$

$\begin{aligned} & \text { let } 2 x+2=\frac{1}{t}-1=\frac{1-t}{t} \\ & I(x)=\int \frac{3\left(-\frac{1}{2 t^2} d t\right)}{2\left(\frac{1}{t}\right) \sqrt{\left(\frac{1-t}{t}\right)^2-1}}=\int \frac{-3 d t}{4 t \sqrt{\frac{1+t^2-2 t-t^2}{t^2}}}=\int \frac{-3 d t}{4 \sqrt{1-2 t}}\end{aligned}$

$\begin{aligned} & I(x)=-\frac{3}{4} \int(1-2 t)^{-1 / 2} d t \\ & I(x)=-\frac{3}{4}\left[\frac{(1-2 t)^{1 / 2}}{(1 / 2)(-2)}\right]+C=\frac{3}{4} \sqrt{1-2 t}+C\end{aligned}$

Substitute $t=\frac{1}{2 x+3}$

$ I(x)=\frac{3}{4} \sqrt{1-\frac{2}{2 x+3}}+C=\frac{3}{4} \sqrt{\frac{2 x+3-2}{2 x+3}}+C=\frac{3}{4} \sqrt{\frac{2 x+1}{2 x+3}}+C $

$I(0)=\frac{\sqrt{3}}{4}+20 \quad$ (Given), substitute $\mathrm{x}=0$ to find value of C.

$\frac{3}{4} \sqrt{\frac{1}{3}}+C=\frac{\sqrt{3}}{4}+20 \Longrightarrow \frac{\sqrt{3}}{4}+C=\frac{\sqrt{3}}{4}+20 \Longrightarrow C=20$

$I(x)=\frac{3}{4} \sqrt{\frac{2 x+1}{2 x+3}}+20$

$ \begin{aligned} x & =\frac{1}{2} \text { for } I\left(\frac{1}{2}\right) \\ I\left(\frac{1}{2}\right) & =\frac{3}{4} \sqrt{\frac{1+1}{1+3}}+20=\frac{3}{4} \sqrt{\frac{2}{4}}+20=\frac{3 \sqrt{2}}{8}+20 \end{aligned} $

Comparing with $\frac{a \sqrt{2}}{b}+c$ :

$ \begin{aligned} & a=3, b=8, c=20 \\ & a+b+c=3+8+20=31 \end{aligned} $

$f(x)=\int \frac{\left(2-x^2\right) \cdot e^x}{(\sqrt{1+x})(1-x)^{3 / 2}} d x$

$\Rightarrow $ $f(x)=\int \frac{\left(2-x^2\right) e^x}{\sqrt{1+x} \sqrt{1-x(1-x)}} d x$

$f(x)=\int \frac{\left(2-x^2\right) \cdot e^x}{\left(\sqrt{1-x^2}\right)(1-x)} d x$

$\begin{aligned} & =\int \frac{\left[1+\left(1-x^2\right)\right] e^x}{\left(\sqrt{1-x^2}\right)(1-x)} d x \\ & =\int e^x\left[\frac{1}{\left(\sqrt{1-x^2}\right)(1-x)}+\frac{1+x}{\sqrt{1-x^2}}\right] d x\end{aligned}$

This integral of the form $\int e^x\left[g(x)+g^{\prime}(x)\right] d x=e^x \cdot g(x)+C$

Let $g(x)=\frac{1+x}{\sqrt{1-x^2}}$

$g^{\prime}(x)=\frac{(1+x)^{\prime} \cdot \sqrt{1-x^2}-(1+x)\left(\sqrt{1-x^2}\right)^{\prime}}{\left(\sqrt{1-x^2}\right)^2}$

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

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

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

$ =\frac{1}{\left(\sqrt{1-x^2}\right)(1-x)} $

So value of integral

$ f(x)=\frac{e^x \cdot(1+x)}{\sqrt{1-x^2}}+C $

Apply initial condition $f(0)=0$ to find $C$

$ 0=\frac{e^0(1+0)}{\sqrt{1-0}}+C \Rightarrow C=-1 $

$\Rightarrow $ $f(x)=\frac{e^x(1+x)}{\sqrt{1-x^2}}-1$

$\therefore $ $f(1 / 2)=\frac{e^{1 / 2}(1+1 / 2)}{\sqrt{1-(1 / 2)^2}}-1$

$=\frac{\sqrt{e} \cdot \frac{3}{2}}{\frac{\sqrt{3}}{2}}-1=\sqrt{3 e}-1$

$\begin{aligned} & \int x^3 \sqrt{3-x^2} d x \\ & 3-x^2=t^2 \\ & -2 x d x=2 t d t \\ & =-\int t^2\left(3-t^2\right) d t=\int t^4-3 t^2 d t \\ & =\frac{t^5}{5}-t^3+c \end{aligned}$

$\begin{aligned} & f(x)=\frac{\left(3-x^2\right)^{\frac{5}{2}}}{5}-\left(3-x^2\right)^{\frac{3}{2}}+c \\ & \because f(\sqrt{2})=-\frac{4}{5} \\ & \Rightarrow \quad-\frac{4}{5}=\frac{1}{5}-1+c \Rightarrow c=0 \\ & \begin{aligned} \therefore \quad f(x) & =\frac{\left(3-x^2\right)^{\frac{5}{2}}}{5}-\left(3-x^2\right)^{\frac{3}{2}} \\ \text { Now } f(1) & =\frac{2^{\frac{5}{2}}}{5}-2^{\frac{3}{2}}=2^{\frac{3}{2}}\left[\frac{2}{5}-1\right] \\ & =2 \sqrt{2}\left(-\frac{3}{5}\right)=-\frac{6}{5} \sqrt{2} \end{aligned} \end{aligned}$

Substitution: Start by substituting $ x = t^4 $ which implies $ dx = 4t^3 \, dt $.

Integral Transformation:

$ \int \frac{1}{x^{1/4}(1 + x^{1/4})} \, dx \Rightarrow \int \frac{4t^3 \, dt}{t(1 + t)} = \int \frac{4t}{1 + t} \, dt $

Rewriting the Integral: The expression $ \frac{4t}{1 + t} $ can be decomposed as:

$ \frac{4t}{1 + t} = 4\left(\frac{t^2 - 1 + 1}{1 + t}\right) = 4\left((t - 1) + \frac{1}{t + 1}\right) $

Separate and Integrate:

$ 4 \int (t - 1) \, dt + 4 \int \frac{1}{t + 1} \, dt $

Solving the Integrals:

The integral of $ t - 1 $ is $ \frac{(t - 1)^2}{2} $.

The integral of $ \frac{1}{t + 1} $ is $ \ln(t + 1) $.

Therefore:

$ f(x) = 4 \left\{ \frac{(t - 1)^2}{2} + \ln(t + 1) \right\} + C $

Substitute $ t = x^{1/4} $: Replace $ t $ back with $ x^{1/4} $:

$ f(x) = 2(x^{1/4} - 1)^2 + 4 \ln(1 + x^{1/4}) + C $

Using the Condition $ f(0) = -6 $:

$ f(0) = 2(0 - 1)^2 + 4 \ln(1 + 0) + C = 2 \times 1 - 0 + C = -6 $

Solving gives $ C = -8 $.

Find $ f(1) $:

$ f(1) = 2(1^{1/4} - 1)^2 + 4 \ln(1 + 1^{1/4}) - 8 $

Simplifies to:

$ f(1) = 0 + 4 \ln 2 - 8 = 4(\ln 2 - 2) $

$\begin{aligned} & \int x^3 \sin x d x=-x^3 \cos x+\int 3 x^2 \cos x d x \\ & =-x^3 \cos x+3 x^2 \sin x-\int 6 x \sin x d x \\ & =-x^3 \cos x+3 x^2 \sin x+6 x \cos x-6 \sin x+c \end{aligned}$

So $g(x)=-x^3 \cos x+3 x^2 \sin x+6 x \cos x-6 \sin x$

$\begin{aligned} & g\left(\frac{\pi}{2}\right)=\frac{3 \pi^2}{4}-6 \\ & g^{\prime}(x)=x^3 \sin x \\ & g^{\prime}\left(\frac{\pi}{2}\right)=\frac{\pi^3}{8} \\ & 8\left(g\left(\frac{\pi}{2}\right)+g^{\prime}\left(\frac{\pi}{2}\right)\right)=\pi^3+6 \pi^2-48 \end{aligned}$

So $\alpha+\beta-\gamma=55$

$\begin{aligned} & I(x)=\int \frac{d x}{(x-11)^{\frac{11}{13}}(x+15)^{\frac{15}{13}}} \\ & \text { Put } \frac{x-11}{x+15}=t \Rightarrow \frac{26}{(x+5)^2} d x=d t \\ & I(x)=\frac{1}{26} \int \frac{d t}{t^{11 / 13}}=\frac{1}{26} \cdot \frac{t^{2 / 13}}{2 / 13} \end{aligned}$

$\begin{aligned} & \mathrm{I}(\mathrm{x})=\frac{1}{4}\left(\frac{\mathrm{x}-11}{\mathrm{x}+15}\right)^{2 / 13}+\mathrm{C} \\ & \mathrm{I}(37)-\mathrm{I}(24)=\frac{1}{4}\left(\frac{26}{52}\right)^{2 / 13}-\frac{1}{4}\left(\frac{13}{39}\right)^{2 / 13} \\ & =\frac{1}{4}\left(\frac{1}{2^{2 / 13}}-\frac{1}{3^{2 / 13}}\right) \\ & =\frac{1}{4}\left(\frac{1}{4^{1 / 13}}-\frac{1}{9^{1 / 13}}\right) \\ & \therefore \mathrm{b}=4, \mathrm{c}=9 \\ & 3(\mathrm{~b}+\mathrm{c})=39 \end{aligned}$

$\begin{aligned} &\begin{aligned} & \because \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{x} \sin ^{-1} \mathrm{x}}{\sqrt{1-\mathrm{x}^2}}\right)=\frac{\sin ^{-1} \mathrm{x}}{\left(1-\mathrm{x}^2\right)^{3 / 2}}+\frac{\mathrm{x}}{1-\mathrm{x}^2} \\ & \Rightarrow \int \mathrm{e}^{\mathrm{x}}\left(\frac{\mathrm{x} \sin ^{-1} \mathrm{x}}{\sqrt{1-\mathrm{x}^2}}+\frac{\sin ^{-1} \mathrm{x}}{\left(1-\mathrm{x}^2\right)^{3 / 2}}+\frac{\mathrm{x}}{1-\mathrm{x}^2}\right) \mathrm{dx} \\ & =\mathrm{e}^{\mathrm{x}} \cdot \frac{\mathrm{x} \sin ^{-1} \mathrm{x}}{\sqrt{1-\mathrm{x}^2}}+\mathrm{c}=\mathrm{g}(\mathrm{x})+\mathrm{C} \end{aligned}\\ &\text { Note : assuming } g(x)=\frac{\mathrm{xe}^{\mathrm{x}} \sin ^{-1} x}{\sqrt{1-\mathrm{x}^2}}\\ &g(1 / 2)=\frac{\mathrm{e}^{1 / 2}}{2} \cdot \frac{\frac{\pi}{6} \times 2}{\sqrt{3}}=\frac{\pi}{6} \sqrt{\frac{\mathrm{e}}{3}} \end{aligned}$

Comment : In this question we will not get a unique function $\mathrm{g}(\mathrm{x})$, but in order to match the answer we will have to assume $g(x)=\frac{\mathrm{xe}^{\mathrm{x}} \sin ^{-1} \mathrm{x}}{\sqrt{1-\mathrm{x}^2}}$.

$I=\int \frac{2-\tan x}{3+\tan x} d x=\int \frac{2 \cos x-\sin x}{3 \cos x+\sin x} d x$

Put, $2 \cos x-\sin x=a(-3 \sin x+\cos x)+ b(3 \cos x+\sin x)$

$\begin{aligned} & a+3 b=2 \quad \text{.... (i)}\\ & -3 a+b=-1 \quad \text{.... (ii)} \end{aligned}$

From equation (i) and (ii) $a=b=\frac{1}{2}$

$I=\frac{1}{2} \int \frac{-3 \sin x+\cos x}{3 \cos x+\sin x} d x+\frac{1}{2} \int \frac{3 \cos x+\sin x}{3 \cos x+\sin x} d x$

$I=\frac{1}{2} \ln |3 \cos x+\sin x|+\frac{1}{2} x= \frac{1}{2}(d x+\log |\beta \sin x+\gamma \cos x|)$

On comparing, we get

$I=\frac{1}{2} \ln |3 \cos x+\sin x|+\frac{1}{2} x=\frac{1}{2}(d x+\log |\beta \sin x+\gamma \cos x|)$

On comparing, we get

$\begin{aligned} & \Rightarrow \frac{1}{2}(x+\log (|3 \cos x+\sin x|))= \\ & \qquad \frac{1}{2}(d x+\log |\beta \sin x+\gamma \cos x|) \end{aligned}$

$\begin{aligned} & \alpha=1, \beta=1, \gamma=3 \\ & \alpha+\frac{\gamma}{\beta}=1+\frac{3}{1}=4 \end{aligned}$

$\begin{aligned} & I(x)=\int \frac{6}{\sin ^2 x(1-\cot x)^2} d x \\ & I(x)=\int \frac{6}{(\sin x-\cos x)^2} d x \\ & =\int \frac{6 \sec ^2 x}{(\tan x-1)^2} d x \end{aligned}$

$\begin{aligned} & \text { Let } \tan x=t \Rightarrow \sec ^2 x d x=d t \\ & =\int \frac{6 d t}{(t-1)^2} \\ & =-\frac{6}{(t-1)}+c \\ & =\frac{-6}{(\tan x-1)}+c \\ & I(x)=\frac{6}{1-\tan x}+c \\ & I(0)=3 \\ & \frac{6}{1-\tan 0}+c=3 \\ & c=-3 \end{aligned}$

$\begin{aligned} I(x) & =\frac{6}{1-\tan x}-3 \\ I\left(\frac{\pi}{12}\right) & =\frac{6}{1-\tan \left(\frac{\pi}{12}\right)}-3 \\ & =\frac{6}{1-(2-\sqrt{3})}-3 \\ & =\frac{6}{\sqrt{3}-1}-3 \\ & =\frac{6-3 \sqrt{3}+3}{\sqrt{3}-1} \\ & =\frac{9-3 \sqrt{3}}{\sqrt{3}-1} \\ & =\frac{3 \sqrt{3}(\sqrt{3}-1)}{\sqrt{3}-1} \\ & =3 \sqrt{3} \end{aligned}$

$\begin{aligned} & \int \frac{1}{a^2 \sin ^2 x+b^2 \cos ^2 x} d x=\frac{1}{12} \tan ^{-1}(3 \tan x)+c \\ & I=\int \frac{\sec ^2 x}{b^2+a^2 \tan ^2 x} d x \\ & \tan x=t \\ & \Rightarrow \sec ^2 x d x=d t \\ & I=\int \frac{d t}{b^2+a^2 t^2} \\ & =\frac{1}{b a} \tan ^{-1}\left(\frac{a t}{b}\right)+c \\ & I=\frac{1}{a b} \tan ^{-1}\left(\frac{a}{b} \tan x\right)+c \\ & \Rightarrow a b=12 \text { and } \frac{a}{b}=3\\ & \Rightarrow a^2=36 \text { and } b^2=4\\ & \Rightarrow \text { Maximum value of } a \sin x+b \cos x \text { is } \sqrt{a^2+b^2} \end{aligned}$

$\quad=\sqrt{36+4}$

$\quad=\sqrt{40}$

$\begin{aligned} & \text {} \int \frac{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}{\sqrt{\sin ^3 x \cos ^3 x \sin (x-\theta)}} d x \\ & I=\int \frac{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}{\sqrt{\sin ^3 x \cos ^3 x(\sin x \cos \theta-\cos x \sin \theta)}} d x \\ & =\int \frac{\sin ^{\frac{3}{2}} x}{\sin ^{\frac{3}{2}} x \cos ^2 x \sqrt{\tan x \cos \theta-\sin \theta}} d x+\int \frac{\cos ^{\frac{3}{2}} x}{\sin ^2 x \cos ^{\frac{3}{2}} x \sqrt{\cos \theta-\cot x \sin \theta}} d x= \\ & \int \frac{\sec ^2 x}{\sqrt{\tan x \cos \theta-\sin \theta}} d x+\int \frac{\operatorname{cosec}^2 x}{\sqrt{\cos \theta-\cot x \sin \theta}} d x \\ & \end{aligned}$

$\mathrm{I}=\mathrm{I}_1+\mathrm{I}_2$ ...... {Let}

For $\mathrm{I}_1$, let $\tan \mathrm{x} \cos \theta-\sin \theta=\mathrm{t}^2$

$\sec ^2 x d x=\frac{2 t d t}{\cos \theta}$

For $\mathrm{I}_2$, let $\cos \theta-\cot \mathrm{x} \sin \theta=\mathrm{z}^2$

$\operatorname{cosec}^2 x d x=\frac{2 z d z}{\sin \theta}$

$\begin{aligned} & I=I_1+I_2 \\ & =\int \frac{2 t d t}{\cos \theta t}+\int \frac{2 z d z}{\sin \theta z} \\ & =\frac{2 t}{\cos \theta}+\frac{2 z}{\sin \theta} \\ & =2 \sec \theta \sqrt{\tan x \cos \theta-\sin \theta}+2 \operatorname{cosec} \theta \sqrt{\cos \theta-\cot x \sin \theta} \\ & \text { Comparing } \\ & \quad A B=8 \operatorname{cosec} 2 \theta \end{aligned}$

$\begin{aligned} & y(x)=\int \frac{\left(1+\sin ^2 x\right) \cos x}{1+\sin ^4 x} d x \\ & \text { Put } \sin x=t \\ & =\int \frac{1+t^2}{t^4+1} d t=\frac{1}{\sqrt{2}} \tan ^{-1} \frac{\left(t-\frac{1}{t}\right)}{\sqrt{2}}+C \\ & x=\frac{\pi}{2}, t=1 \quad \therefore C=0 \\ & y\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}} \tan ^{-1}\left(-\frac{1}{2}\right) \end{aligned}$

$I=\int \frac{x^8-x^2}{\left(x^{12}+3 x^6+1\right) \tan ^{-1}\left(x^3+\frac{1}{x^3}\right)} d x$

$\begin{aligned} & \text { Let } \tan ^{-1}\left(x^3+\frac{1}{x^3}\right)=t \\ & \Rightarrow \frac{1}{1+\left(x^3+\frac{1}{x^3}\right)^2} \cdot\left(3 x^2-\frac{3}{x^4}\right) d x=d t \\ & \Rightarrow \frac{x^6}{x^{12}+3 x^6+1} \cdot \frac{3 x^6-3}{x^4} d x=d t \\ & I=\frac{1}{3} \int \frac{d t}{t}=\frac{1}{3} \ln |t|+C \\ & I=\frac{1}{3} \ln \left|\tan ^{-1}\left(x^3+\frac{1}{x^3}\right)\right|+C \\ & I=\ln \left|\tan ^{-1}\left(x^3+\frac{1}{x^3}\right)\right|^{1 / 3}+C \end{aligned}$

Hence option (1) is correct.

To solve the integral:

$ I = \int\left[\left(\frac{x}{2}\right)^x+\left(\frac{2}{x}\right)^x\right] \ln \left(\frac{e x}{2}\right) \, dx $

we start by simplifying the logarithmic term:

$ \ln\left(\frac{e x}{2}\right) = \ln(e) + \ln\left(\frac{x}{2}\right) = 1 + \ln\left(\frac{x}{2}\right) $

So the integral becomes:

$ I = \int \left[ \left( \frac{x}{2} \right)^x + \left( \frac{2}{x} \right)^x \right] \left[ 1 + \ln\left( \frac{x}{2} \right) \right] \, dx $

Let's denote:

$ A = \left( \dfrac{x}{2} \right)^x $

$ B = \left( \dfrac{2}{x} \right)^x $

Then, the integral can be split:

$ I = \int \left[ A(1 + \ln\left( \frac{x}{2} \right)) + B(1 + \ln\left( \frac{x}{2} \right)) \right] \, dx $

Calculating $ \int A(1 + \ln\left( \frac{x}{2} \right)) \, dx $:

First, find the derivative of $ A $:

$ \frac{dA}{dx} = A \left[ \ln\left( \frac{x}{2} \right) + 1 \right] $

This means:

$ A \left[ 1 + \ln\left( \frac{x}{2} \right) \right] = \frac{dA}{dx} $

Therefore:

$ \int A \left[ 1 + \ln\left( \frac{x}{2} \right) \right] \, dx = \int \frac{dA}{dx} \, dx = A + C_1 = \left( \frac{x}{2} \right)^x + C_1 $

Calculating $ \int B(1 + \ln\left( \frac{x}{2} \right)) \, dx $:

Note that:

$ \ln\left( \frac{x}{2} \right) = \ln x - \ln 2 $

$ \ln\left( \frac{2}{x} \right) = \ln 2 - \ln x $

The derivative of $ B $ is:

$ \frac{dB}{dx} = B \left[ \ln\left( \frac{2}{x} \right) + 1 \right] = B \left[ \ln 2 - \ln x + 1 \right] $

But in our integrand, we have $ B \left[ 1 + \ln\left( \frac{x}{2} \right) \right] = B \left[ 1 + \ln x - \ln 2 \right] $. Notice that:

$ 1 + \ln x - \ln 2 = - \left( \ln 2 - \ln x + 1 \right) $

Thus:

$ B \left[ 1 + \ln\left( \frac{x}{2} \right) \right] = - \frac{dB}{dx} $

Therefore:

$ \int B \left[ 1 + \ln\left( \frac{x}{2} \right) \right] \, dx = -\int \frac{dB}{dx} \, dx = -B + C_2 = -\left( \frac{2}{x} \right)^x + C_2 $

Combining the Results:

Adding both integrals:

$ I = \left( \frac{x}{2} \right)^x - \left( \frac{2}{x} \right)^x + C $

Answer: Option B

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

Given,

$I(x) = \int {{{{{\sec }^2}x - 2022} \over {{{\sin }^{2022}}x}}dx} $

$ = \int {{{{{\sec }^2}x} \over {{{\sin }^{2022}}x}}dx - \int {{{2022} \over {{{\sin }^{2022}}x}}dx} } $

$ = \int {{1 \over {{{\sin }^{2022}}x}}\,.\,{{\sec }^2}x\,dx - \int {{{2022} \over {{{\sin }^{2022}}x}}dx} } $

$ = {1 \over {{{\sin }^{2022}}x}}\,.\,\tan x - \int {\left( {{{ - 2022} \over {{{\sin }^{2023}}x}}\,.\,\cos x\,.\,\tan x} \right)dx - \int {{{2022} \over {{{\sin }^{2022}}x}}dx + C} } $

$ = {{\tan x} \over {{{\sin }^{2022}}x}} + \int {\left( {{{2022} \over {{{\sin }^{2023}}x}}\,.\,\cos x\,.\,{{\sin x} \over {\cos x}}} \right)dx - \int {{{2022} \over {{{\sin }^{2022}}x}}dx + C} } $

$ = {{\tan x} \over {{{\sin }^{2022}}x}} + \int {{{2022} \over {{{\sin }^{2022}}x}}dx - \int {{{2022} \over {{{\sin }^{2022}}x}}dx} } $

$ = {{\tan x} \over {{{\sin }^{2022}}x}} + C$

Given, $I\left( {{\pi \over 4}} \right) = {2^{1011}}$

$\therefore$ $I\left( {{\pi \over 4}} \right) = {{\tan \left( {{\pi \over 4}} \right)} \over {{{\left( {\sin {\pi \over 4}} \right)}^{2022}}}} + C$

$ \Rightarrow {2^{1011}} = {1 \over {{{\left( {{1 \over {\sqrt 2 }}} \right)}^{2022}}}} + C$

$ \Rightarrow C = {2^{1011}} - {2^{1011}} = 0$

$\therefore$ $I(x) = {{\tan x} \over {{{\sin }^{2022}}x}}$

$\therefore$ $I\left( {{\pi \over 3}} \right) = {{\tan {\pi \over 3}} \over {{{\left( {\sin {\pi \over 3}} \right)}^{2022}}}} = {{\sqrt 3 } \over {{{\left( {{{\sqrt 3 } \over 2}} \right)}^{2022}}}}$

$I\left( {{\pi \over 6}} \right) = {{{1 \over {\sqrt 3 }}} \over {{{\left( {{1 \over 2}} \right)}^{2022}}}} = {1 \over {\sqrt 3 }} \times {(2)^{2022}}$

From option (A),

${3^{1010}}\,.\,I\left( {{\pi \over 3}} \right) - I\left( {{\pi \over 6}} \right)$

$ = {3^{1010}}\,.\,\sqrt 3 \,.\,{\left( {{2 \over {\sqrt 3 }}} \right)^{2022}} - {{{{(2)}^{2022}}} \over {\sqrt 3 }}$

$ = {3^{1010}}\,.\,\sqrt 3 \times {{{2^{2022}}} \over {{3^{1011}}}} - {{{2^{2022}}} \over {\sqrt 3 }}$

$ = {{{2^{2022}}} \over {\sqrt 3 }} - {{{2^{2022}}} \over {\sqrt 3 }} = 0$

$ = \int {{{\left( {1 - {1 \over {\sqrt 3 }}} \right)(\cos x - \sin x)} \over {\left( {1 + {2 \over {\sqrt 3 }}\sin 2x} \right)}}dx} $

$ = \int {{{\left( {{{\sqrt 3 - 1} \over {\sqrt 3 }}} \right)\sqrt 2 \sin \left( {{\pi \over 4} - x} \right)} \over {\left( {{2 \over {\sqrt 3 }}} \right)\left( {\sin {\pi \over 3} + \sin 2x} \right)}}dx} $

$ = \int {{{{{(\sqrt 3 - 1)} \over {\sqrt 2 }}\sin \left( {{\pi \over 4} - x} \right)} \over {\left( {\sin {\pi \over 3} + \sin 2x} \right)}}dx} $

$ = \int {{{{{\sqrt 3 - 1} \over {2\sqrt 2 }}\sin \left( {{\pi \over 4} - x} \right)} \over {\sin \left( {{\pi \over 6} + x} \right)\cos \left( {{\pi \over 6} - x} \right)}}dx} $

$ = {1 \over 2}\int {{{2\sin {\pi \over {12}}\sin \left( {{\pi \over 4} - x} \right)} \over {\sin \left( {{\pi \over 6} + x} \right)\cos \left( {{\pi \over 6} - x} \right)}}dx} $

$ = {1 \over 2}\int {{{\cos \left( {{\pi \over 6} - x} \right) - \cos \left( {{\pi \over 3} - x} \right)} \over {\sin \left( {{\pi \over 6} + x} \right)\cos \left( {{\pi \over 6} - x} \right)}}dx} $

$ = {1 \over 2}\left[ {\int {{\mathop{\rm cosec}\nolimits} \left( {{\pi \over 6} + x} \right)dx - \int {\sec \left( {{\pi \over 6} - x} \right)dx} } } \right]$

$ = {1 \over 2}\left[ {\ln \left| {\tan \left( {{\pi \over {12}} + {x \over 2}} \right)} \right| - \int {\cos ec\left( {{\pi \over 3} - x} \right)dx} } \right]$

$ = {1 \over 2}\left[ {\ln \left| {\tan \left( {{\pi \over {12}} + {x \over 2}} \right)} \right| - \ln \left| {{\pi \over 6} + {x \over 2}} \right|} \right] + C$

$ = {1 \over 2}\ln \left| {{{\tan \left( {{\pi \over 2} + {x \over 2}} \right)} \over {\tan \left( {{\pi \over 6} + {x \over 2}} \right)}}} \right| + C$

$I = \int {{{{e^x}({x^2} + 1)} \over {{{(x + 1)}^2}}}dx = f(x){e^x} + c} $

$ = \int {{{{e^x}({x^2} - 1 + 1 + 1)} \over {{{(x + 1)}^2}}}dx} $

$ = \int {{e^x}\left[ {{{x - 1} \over {x + 1}} + {2 \over {{{(x + 1)}^2}}}} \right]dx} $

$ = {e^x}\left( {{{x - 1} \over {x + 1}}} \right) + c$

$\therefore$ $f(x) = {{x - 1} \over {x + 1}}$

$f(x) = 1 - {2 \over {x + 1}}$

$f'(x) = 2{\left( {{1 \over {x + 1}}} \right)^2}$

$f''(x) = - 4{\left( {{1 \over {x + 1}}} \right)^3}$

$f'''(x) = {{12} \over {{{(x + 1)}^4}}}$

for $x = 1$

$f'''(1) = {{12} \over {{2^4}}} = {{12} \over {16}} = {3 \over 4}$

To solve the given integral, first we simplify the expression in the integral as follows :

$\int {{{\tan x + \tan \alpha } \over {\tan x - \tan \alpha }}} dx = \int {{{\sin x \over \cos x} + {\sin \alpha \over \cos \alpha }} \over {{\sin x \over \cos x} - {\sin \alpha \over \cos \alpha }}} dx$

Simplifying further, this becomes :

$\int {{{\sin x\cos \alpha + \sin \alpha \cos x} \over {\sin x\cos \alpha - \sin \alpha \cos x}}} dx$

By using the formula for sin(a + b) = sin a cos b + cos a sin b and sin(a - b) = sin a cos b - cos a sin b, we get :

$\int {{{\sin(x + \alpha)}} \over {\sin(x - \alpha)}} dx$

Now, let's use the substitution method. Let t = x - α, therefore x = y + α and dx = dt. Substituting these values into the integral, we get :

= $\int {{{\sin(t + 2\alpha)}} \over {\sin y}} dy$