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Question

$\int\limits^\pi_0\frac{\text{x}}{1+\sin\text{x}}$

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Answer

Let $\text{I}=\int\limits^\pi_0\frac{\text{x}}{1+\sin\text{x}}\text{dx}\ \ \dots(\text{i})$
and $\text{I}=\int\limits^\pi_0\frac{\pi-\text{x}}{1+\sin(\pi-\text{x})}\text{dx}$ $=\text{I}=\int\limits^\pi_0\frac{\pi-\text{x}}{1+\sin\text{x}}\text{dx}\ \ \dots(\text{ii})$
On adding Eqs. (i) and (ii), we get
$2\text{I}\int\limits^\pi_0\frac{1}{1+\sin\text{x}}\text{dx}$
$=\pi\int\limits^\pi_0\frac{(1-\sin\text{x})\text{dx}}{(1+\sin\text{x})(1-\sin\text{x})}$
$=\pi\int\limits^\pi_0\frac{(1-\sin\text{x})\text{dx}}{\cos^2\text{x}}$
$=\pi\int\limits^\pi_0\big(\sec^2\text{x}-\tan\text{x}.\sec\text{x}\big)\text{dx}$
$=\pi\int\limits^\pi_0\sec^2\text{x dx}-\pi\int\limits^\pi_0\sec\text{x}\tan\text{x dx}$
$=\pi(\tan\text{x})^\pi_0-(\sec\text{x})^\pi_0$
$=\pi(\tan\pi-\tan0-\sec\pi+\sec0)$
$\Rightarrow2\text{I}=\pi(0-0+1+1)=2\pi$
$2\text{I}=2\pi$
$\therefore\ \text{I}=\pi$

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