Definite Integration — Maths STD 12 Science — Question

 (c) : Let $\tan ^{-1} x=\theta \Rightarrow x=\tan \theta$
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\Rightarrow d x=\sec ^2 \theta d \theta
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Now, $x=0 \Rightarrow \theta=0$ and $x=1 \Rightarrow \theta=\frac{\pi}{4}$
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\begin{aligned}
& \therefore \quad \int_0^1 \frac{x \tan ^{-1} x}{\left(1+x^2\right)^{3 / 2}} d x=\int_0^{\pi / 4} \frac{\theta \tan \theta}{\sec ^3 \theta} \sec ^2 \theta d \theta \\
& =\int_0^{\pi / 4} \theta \sin \theta d \theta=[-\theta \cos \theta]_0^{\pi / 4}-\int_0^{\pi / 4}(-\cos \theta) d \theta \\
&
\end{aligned}
$
$=[-\theta \cos \theta]_0^{\pi / 4}+[\sin \theta]_0^{\pi / 4}=\frac{4-\pi}{4 \sqrt{2}}$