Let [I(x)= x^2(x ^2 x+ x)(x x+1)^2 d x].

Question

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,

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

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

Now, let

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

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

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

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

When, $x=0$, then

$$
\[\begin{gathered} I(0)=0+2 \log (1)+c=0 \ \Rightarrow c=0 \ \therefore I(x)=\frac{-x^2}{x \tan x+1}+2 \log (x \sin x+\cos x) \end{gathered}$$
\]

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

Solutions · 1