Question
Evaluate the following integrals:
$\int^\limits\frac{\pi}{2}_{0}\frac{\cos^2\text{x}}{1+3\sin^3\text{x}}\text{ dx}$
$\int^\limits\frac{\pi}{2}_{0}\frac{\cos^2\text{x}}{1+3\sin^3\text{x}}\text{ dx}$
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Answer
$\text{I}=\int^\limits\frac{\pi}{2}_{0}\frac{\cos^2\text{x}}{1+3\sin^3\text{x}}\text{ dx}$
$\text{I}=\int^\limits\frac{\pi}{2}_{0}\frac{\sec^2\text{x}}{\sec^2\text{x}\big(\sec^2\text{x}+3\tan^2\text{x}\big)}\text{ dx}$
Put $\tan\text{x}=\text{t}$
$\sec^2\text{x dx}=\text{dt}$
$\text{x}=0\Rightarrow\text{t}=0$ and $\text{x}=\frac{\pi}{2}\Rightarrow\text{t}=\infty$
$\Rightarrow\text{I}=\int^{\infty}\limits_0\frac{1}{\big(1+\text{t}^2\big)\big(1+4\text{t}^2\big)}\text{ dt}$
$\Rightarrow\text{I}=-\frac{1}{3}\int^{\infty}\limits_0\bigg[\frac{1}{(1+\text{t}^2)-{(1+4\text{t}^2)}}\bigg]\text{dt}$
$\Rightarrow\text{I}=-\frac{1}{3}\Big[\tan^{-1}\text{t}-2\tan^{-1}2\text{t}\Big]^{\infty}_0$
$\Rightarrow\text{I}=\frac{\pi}{6}$
$\text{I}=\int^\limits\frac{\pi}{2}_{0}\frac{\sec^2\text{x}}{\sec^2\text{x}\big(\sec^2\text{x}+3\tan^2\text{x}\big)}\text{ dx}$
Put $\tan\text{x}=\text{t}$
$\sec^2\text{x dx}=\text{dt}$
$\text{x}=0\Rightarrow\text{t}=0$ and $\text{x}=\frac{\pi}{2}\Rightarrow\text{t}=\infty$
$\Rightarrow\text{I}=\int^{\infty}\limits_0\frac{1}{\big(1+\text{t}^2\big)\big(1+4\text{t}^2\big)}\text{ dt}$
$\Rightarrow\text{I}=-\frac{1}{3}\int^{\infty}\limits_0\bigg[\frac{1}{(1+\text{t}^2)-{(1+4\text{t}^2)}}\bigg]\text{dt}$
$\Rightarrow\text{I}=-\frac{1}{3}\Big[\tan^{-1}\text{t}-2\tan^{-1}2\text{t}\Big]^{\infty}_0$
$\Rightarrow\text{I}=\frac{\pi}{6}$
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