Evaluate the definite integral in Exercise: ^ 2 _ 0 ^ 2 dx ^ 2 x+… - Vidyadip

Question

Evaluate the definite integral in Exercise:
$\int\limits^{\frac{\pi}{2}}_{0}\frac{\cos^{2}\text{dx}}{\cos^{2}\text{x}+4\sin^{2}\text{x}}$

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Answer

$\text{Let I}=\int^{\frac{\pi}{2}}\limits_{0}\frac{\cos^{2}\text{dx}}{\cos^{2}\text{x}+4\sin^{2}\text{x}}\text{dx}$$\Rightarrow\text{I}=\int^{\frac{\pi}{2}}\limits_{0}\frac{\cos^{2}\text{x}}{\cos^{2}\text{x}+4(1-\cos^{2}\text{x)}}\text{dx}$
$\Rightarrow\text{I}=\int^{\frac{\pi}{2}}\limits_{0}\frac{\cos^{2}\text{x}}{\cos^{2}\text{x}+4-4\cos^{2}\text{x}}\text{dx}$
$\Rightarrow\text{I}=\frac{-1}{3}\int^{\frac{\pi}{2}}\limits_{0}\frac{4-3\cos^{2}\text{x}-4}{4-3\cos^{2}\text{x}}\text{dx}$
$\Rightarrow\text{I}=\frac{-1}{3}\int^{\frac{\pi}{2}}\limits_{0}\frac{4-3\cos^{2}\text{x}}{4-3\cos^{2}\text{x}}\text{dx}+\frac{1}{3}\int^{\frac{\pi}{2}}\limits_{0}\frac{4}{4-3\cos^{2}\text{x}}\text{dx}$
$\Rightarrow\text{I}=\frac{-1}{3}\int^{\frac{\pi}{2}}\limits_{0}1\text{dx}+\frac{1}{3}\int^{\frac{\pi}{2}}\limits_{0}\frac{4\sec^{2}\text{x}}{4\sec^{2}\text{x}-3}\text{dx}$
$\Rightarrow\text{I}=\frac{-1}{3}[\text{x}]^{\frac{\pi}{2}}_{0}+\frac{1}{3}\int^{\frac{\pi}{2}}\limits_{0}\frac{4\sec^{2}\text{x}}{4(1\tan^{2}\text{x})-3}\text{dx}$
$\Rightarrow\text{I}=-\frac{\pi}{6}+\frac{2}{3}\int^{\frac{\pi}{2}}_{0}\frac{2\sec^{2}}{1+4\tan^{2}\text{x}}\text{dx}$
Consider, $\int^{\frac{\pi}{2}}\limits_{0}\frac{2\sec^{2}\text{x}}{1+4\tan^{2}\text{x}}\text{dx}$
$\text{Let} \ 2\tan\text{x}=\text{t}\Rightarrow2\sec^{2}\text{x dx}=\text{dt}$
When $\text{x}=0,\text{t}=0$ and when $\text{x}=\frac{\pi}{2},\text{t} = \infty$
$\Rightarrow\int^{\frac{\pi}{2}}\limits_{0}\frac{2\sec^{2}\text{x}}{1+4\tan^{2}\text{x}}\text{dx}=\int^{\infty}_{0}\frac{\text{dt}}{1+\text{t}^{2}}$
$=\bigg[\tan^{-1}\text{t}\bigg]^{\infty}_{0}$
$=\Big[\tan^{-1}(\infty)-\tan^{-1}(0)\Big]$
$=\frac{\pi}{2}$
Therefore, from (1), we obtain
$\text{I}=-\frac{\pi}{6}+\frac{2}{3}\Big[\frac{\pi}{2}\Big]=\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}$

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