Evaluate the following integrals: ^ _0 ( x 1+ ^2x + ^7x )dx

Question

Evaluate the following integrals:$\int\limits^\pi_0\Big(\frac{\text{x}}{1+\sin^2\text{x}}+\cos^7\text{x}\Big)\text{dx}$

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Answer

Let $\text{I}=\int\limits^\pi_0\Big(\frac{\text{x}}{1+\sin^2\text{x}}+\cos^7\text{x}\Big)\text{dx}\ ....(\text{i})$ Then,$\text{I}=\int\limits^\pi_0\Big(\frac{\pi-\text{x}}{1+\sin^2(\pi-\text{x})}+\cos^7(\pi-\text{x})\Big)\text{dx}$
$=\int\limits^\pi_0\Big(\frac{\pi-\text{x}}{1+\sin^2\text{x}}+\cos^7\text{x}\Big)\text{dx}\ ...(\text{ii})$
Adding (i) and (ii) we get$2\text{I}=\int\limits^\pi_0\Big(\frac{\text{x}}{1+\sin^2\text{x}}+\cos^7\text{x}+\frac{\pi-\text{x}}{1+\sin^2\text{x}}-\cos^7\text{x}\Big)\text{dx}$
$\Rightarrow2\text{I}=\pi\int\limits^\pi_0\frac{1}{1+\sin^2\text{x}}\text{ dx}$
Dividing the numerator and denominator by $\cos^2\text{x},$ we get$2\text{I}=\pi\int\limits^\pi_0\frac{\sec^2\text{x}}{\sec^2\text{x}+\tan^2\text{x}}\text{ dx}$
$\Rightarrow2\text{I}=\pi\int\limits^\pi_0\frac{\sec^2\text{x}}{1+2\tan^2\text{x}}\text{ dx}$
$\Rightarrow2\text{I}=2\pi\int\limits^\pi_0\frac{\sec^2\text{x}}{1+2\tan^2\text{x}}\text{ dx}$ $\begin{bmatrix}\int\limits^{2\text{a}}_0\text{f(x)}\text{dx}=\begin{cases}\int\limits^{2\text{a}}_0\text{f(x)}\text{dx},&\text{if }\text{f}(2\text{a}-\text{x})=\text{f(x)}\\0,&\text{if }\text{f}(2\text{a}-\text{x})=\text{f(x)}\end{cases}\end{bmatrix}$
Put $\tan\text{x}=\text{z}$ Then $\sec^2\text{x dx}=\text{dz}$ When $\text{x}\rightarrow0,\text{ z}\rightarrow0$ When $\text{x}\rightarrow\frac{\pi}{2},\text{ z}\rightarrow\infty$$\therefore\ 2\text{I}=2\pi\int\limits^{\infty}_0\frac{\text{dz}}{1+\big(\sqrt{2}\text{z}\big)^2}$
$\Rightarrow2\text{I}=2\pi\times\bigg[\frac{\tan^{-1}\sqrt{2}\text{z}}{\sqrt{2}}\bigg]^{\infty}_0$
$\Rightarrow\text{I}=\frac{\pi}{\sqrt{2}}\Big(\tan^{-1}\infty-\tan^{-1}0\Big)$
$\Rightarrow\text{I}=\frac{\pi}{\sqrt{2}}\times\Big(\frac{\pi}{2}-0\Big)$
$\Rightarrow\text{I}=\frac{\pi^2}{2\sqrt{2}}$