Evaluate the definite integral in Exercise: ^ _ 0 x + dx - Vidyadip

Question

Evaluate the definite integral in Exercise:
$\int^{\pi}_{0}\frac{\text{x}\tan\text{x}}{\sec\text{x}+\tan\text{x}}\text{dx}$

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Answer

$\text{Let I}=\int^{\pi}_{0}\frac{\text{x}\tan\text{x}}{\sec\text{x}+\tan\text{x}}\text{dx}$
$\text{I}=\int^{\pi}_{0}\left\{\frac{(\pi-\text{x)}\tan(\pi-\text{x)}}{\sec(\pi-\text{x)}+\tan(\pi-\text{x})}\text{dx}\right\}\ \ \Bigg(\int^{\text{a}}\limits_{0}\text{f (x)dx}=\int^{\text{a}}\limits_{0}\text{f (a}-\text{x)dx}\Bigg)$
$\Rightarrow\text{I}=\int^{\pi}\limits_{0}\left\{\frac{-(\pi-\text{x)}\tan\text{x}}{-(\sec\text{x}+\tan\text{x)}}\text{dx}\right\}$
$\Rightarrow\text{I}=\int^{\pi}\limits_{0}\frac{(\pi-\text{x)}\tan\text{x}}{\sec\text{x}+\tan\text{x}}\text{dx}$
Adding (1) and (2), we obtain
$2\text{I}=\int^{\pi}\limits_{0}\frac{{\pi}\tan\text{x}}{\sec\text{x}+\tan\text{x}}\text{dx}$
$\Rightarrow2\text{I}=\pi\int^{\pi}\limits_{0}\frac{\frac{\sin\text{x}}{\cos\text{x}}}{\frac{1}{\cos\text{x}}+\frac{\sin\text{x}}{\cos\text{x}}}\text{dx}$
$\Rightarrow2\text{I}=\pi\int^{\pi}\limits_{0}\frac{\sin\text{x}+1-1}{1+\sin\text{x}}\text{dx}$
$\Rightarrow2\text{I}=\pi\int^{\pi}\limits_{0}1.\text{dx}-\pi\int^{\pi}\limits_{0}\frac{1}{1+\sin\text{x}}\text{dx}$
$\Rightarrow2\text{I}=\pi[\text{x}]^{\pi}_{0}-\pi\int^{\pi}\limits_{0}\frac{1-\sin\text{x}}{\cos^{2}\text{x}}\text{dx}$
$\Rightarrow2\text{I}=\pi^{2}-\pi\int^{\pi}\limits_{0}(\sec^{2}\text{x}-\tan\text{x}\sec\text{x)}\text{dx}$
$\Rightarrow2\text{I}=\pi^{2}-\pi\big[\tan\text{x}-\sec\text{x}\big]^{\pi}_{0}$
$\Rightarrow2\text{I}=\pi^{2}-\pi\big[\tan\pi-\sec\pi-\tan0+\sec0\big]$
$\Rightarrow2\text{I}=\pi^{2}-\pi\big[0-(-1)+1\big]$
$\Rightarrow2\text{I}=\pi^{2}-2\pi$
$\Rightarrow2\text{I}=\pi(\pi-2)$
$\Rightarrow\text{I}=\frac{\pi}{2}(\pi-2)$

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