Evaluate the following integrals: ^ 2 _02 ^ -1 ( )dx - Vidyadip

Question

Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_02\sin\text{x }\cos\text{x}\tan^{-1}(\sin\text{x})\text{dx}$

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Answer

Let $\text{I}=\int^\limits{\frac{\pi}{2}}_02\sin\text{x }\cos\text{x}\tan^{-1}(\sin\text{x})\text{dx}$ Then,
Let $\sin\text{x}=\text{t}$ Then, $\cos\text{x dx}=\text{dt}$
When $\text{x}=0,\text{t}=0$ and $\text{x}=\frac{\pi}{2},\text{t}=1$
$\Rightarrow\text{I}=2\Big[\frac{\text{t}^2\tan^{-1}\text{t}}{2}\Big]^1_0-\int^\limits1_0\frac{\text{t}^2}{1+\text{t}^2}\text{ dt}$
$\Rightarrow\text{I}=2\Big[\frac{\text{t}^2\tan^{-1}\text{t}}{2}\Big]^1_0-\int^\limits1_0\Big(\frac{1+\text{t}^3}{1+\text{t}^2}-\frac{1}{1+\text{t}^2}\Big)\text{dt}$
$\Rightarrow\text{I}=2\Big[\frac{\text{t}^2\tan^{-1}\text{t}}{2}\Big]^1_0-\big[\text{t}-\tan^{-1}\text{t}\big]^1_0$
$\Rightarrow\text{I}=1\tan^{-1}1-0-1+\tan^{-1}1+0$
$\Rightarrow\text{I}=\frac{\pi}{4}-1+\frac{\pi}{4}$
$\Rightarrow\text{I}=\frac{\pi}{2}-1$

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