INTEGRALS — Maths STD 12 Science — Question

$\Rightarrow\text{I}=\int^{\frac{\pi}{2}}\limits_{0}\frac{\cos^{2}\text{x}}{\cos^{2}\text{x}+4(1-\cos^{2}\text{x)}}\text{dx}$

$\Rightarrow\text{I}=\int^{\frac{\pi}{2}}\limits_{0}\frac{\cos^{2}\text{x}}{\cos^{2}\text{x}+4-4\cos^{2}\text{x}}\text{dx}$

$\Rightarrow\text{I}=\frac{-1}{3}\int^{\frac{\pi}{2}}\limits_{0}\frac{4-3\cos^{2}\text{x}-4}{4-3\cos^{2}\text{x}}\text{dx}$

$\Rightarrow\text{I}=\frac{-1}{3}\int^{\frac{\pi}{2}}\limits_{0}\frac{4-3\cos^{2}\text{x}}{4-3\cos^{2}\text{x}}\text{dx}+\frac{1}{3}\int^{\frac{\pi}{2}}\limits_{0}\frac{4}{4-3\cos^{2}\text{x}}\text{dx}$

$\Rightarrow\text{I}=\frac{-1}{3}\int^{\frac{\pi}{2}}\limits_{0}1\text{dx}+\frac{1}{3}\int^{\frac{\pi}{2}}\limits_{0}\frac{4\sec^{2}\text{x}}{4\sec^{2}\text{x}-3}\text{dx}$

$\Rightarrow\text{I}=\frac{-1}{3}[\text{x}]^{\frac{\pi}{2}}_{0}+\frac{1}{3}\int^{\frac{\pi}{2}}\limits_{0}\frac{4\sec^{2}\text{x}}{4(1\tan^{2}\text{x})-3}\text{dx}$

$\Rightarrow\text{I}=-\frac{\pi}{6}+\frac{2}{3}\int^{\frac{\pi}{2}}_{0}\frac{2\sec^{2}}{1+4\tan^{2}\text{x}}\text{dx}$

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

$\text{I}=-\frac{\pi}{6}+\frac{2}{3}\Big[\frac{\pi}{2}\Big]=\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}$