QJEEM23Maf3be: JEE Main 2023 Mathematics - Class 12 JEEM23Maf3be Solution: Let . If , then is equal to :

Question

Verified Solution • 2025–26 Session moderate

Let $I(x)=\int \frac{x^{2}\left(x \sec ^{2} x+\tan x\right)}{(x \tan x+1)^{2}} d x$ . If $I(0)=0$ , then $I\left(\frac{\pi}{4}\right)$ is equal to :

Accepted Answer

We have,

\begin{aligned}I(x)= &amp; \int \frac{x^2\left(x \sec ^2 x+\tan x\right)}{(x \tan x+1)^2} d x \\\\= &amp; x^2 \int \frac{x \sec ^2 x+\tan x}{(x \tan x+1)^2} d x \\\\&amp; \quad-\int\left\{\frac{d}{d x}\left(x^2\right) \int \frac{x \sec ^2 x+\tan x}{(x \tan x+1)^2} d x\right\} d x \text { (integration by parts) }\end{aligned}

=x^2\left(\frac{-1}{x \tan x+1}\right)+\int \frac{2 x}{x \tan x+1} d x

Now, let

\begin{aligned}I_1 &amp; =2 \int \frac{x}{x \tan x+1} d x \\\\&amp; =2 \int \frac{x \cos x}{x \sin x+\cos x} d x\end{aligned}

\begin{aligned}&amp; \text { On putting } x \sin x+\cos x=t \\\\&amp; \Rightarrow (x \cos x+\sin x-\sin x) d x=d t \\\\&amp; \Rightarrow x \cos x d x=d t\end{aligned}

\begin{aligned}&amp;\therefore I_1 =2 \int \frac{d t}{t}=2 \log t+c \\\\&amp; =2 \log (x \sin x+\cos x)+c\end{aligned}

\begin{gathered}\therefore I(x)=\frac{-x^2}{x \tan x+1}+2 \log(x \sin x+\cos x)+c\end{gathered}

When,

x=0

, then

\begin{array}{ll} &amp; I(0)=0+2 \log (1)+c=0 \\\\&amp;\Rightarrow c=0 \\\\&amp;\therefore I(x)=\frac{-x^2}{x \tan x+1}+2 \log (x \sin x+\cos x)\end{array}

\begin{aligned}&amp;\therefore I(x) =\frac{-x^2}{x \tan x+1}+2 \log (x \sin x+\cos x) \\\\&amp;\Rightarrow I\left(\frac{\pi}{4}\right) =\frac{-\frac{\pi^2}{16}}{\frac{\pi}{4}+1}+2 \log \left(\frac{\pi}{4} \times \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right) \\\\&amp; =\log \left(\frac{(\pi+4)^2}{32}\right)-\frac{\pi^2}{4(\pi+4)}\end{aligned}

Step-by-Step Verified Solution