Evaluate the following integrals: ^ _ - 2x(1+ ) 1+ ^2x dx

Question

Evaluate the following integrals:
$\int\limits^{{\pi}}_{{-\pi}}\frac{2\text{x}(1+\sin\text{x})}{1+\cos^2\text{x}}\text{ dx}$

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Answer

Let $\text{I}=\int\limits^{{\pi}}_{{-\pi}}\frac{2\text{x}(1+\sin\text{x})}{1+\cos^2\text{x}}\text{ dx}$
Then,
$\text{I}=\int\limits^{{\pi}}_{{-\pi}}\frac{2\text{x}}{1+\cos^2\text{x}}\text{ dx}+\int\limits^{{\pi}}_{{-\pi}}\frac{2\text{x}\sin\text{x}}{1+\cos^2\text{x}}\text{ dx}$
$=\text{I}_1+\text{I}_2$
Consider $\text{f(x)}=\frac{2\text{x}}{1+\cos^2-\text{x}}$
Now,
$\text{f}(-\text{x})=\frac{2(-\text{x})}{1+\cos^2(\pi-\text{x})}=-\frac{2\text{x}}{1+(-\cos\text{x})^2}=-\frac{2\text{x}}{1+\cos^2\text{x}}=-\text{f(x)}$
$\therefore\ \text{I}_1=\int\limits^{{\pi}}_{{-\pi}}\frac{2\text{x}}{1+\cos^2\text{x}}\text{ dx}=0$ $\begin{bmatrix}\int\limits^{\text{a}}_{-\text{a}}\text{f(x)}\text{dx}=\begin{cases}2\int\limits^{\text{a}}_{0}\text{f(x)}\text{dx},&\text{ if }\text{f}(-\text{x})=\text{f(x)}\\0,&\text{ if }\text{f}(-\text{x})=-\text{f}(\text{x})\end{cases}\end{bmatrix}$
Again, consider $\text{g(x)}=\frac{2\text{x}\sin\text{x}}{1+\cos^2\text{x}}$
$\text{g}(-\text{x})=\frac{2(-\text{x})\sin(-\text{x})}{1+\cos^2(-\text{x})}=\frac{2\text{x}\sin\text{x}}{1+\cos^2\text{x}}=\text{g(x)}$ $\big[\sin(-\text{x})=-\sin\text{x}\text{ and }\cos(-\text{x})=\cos\text{x}\big]$
$\therefore\ \text{I}_2=\int\limits^{\pi}_{-\pi}\frac{2\text{x}\sin\text{x}}{1+\cos^2\text{x}}\text{ dx}$
$=2\times2\int\limits^{\pi}_0\frac{\text{x}\sin\text{x}}{1+\cos^2\text{x}}\text{ dx}$ $\begin{bmatrix}\int\limits^{\text{a}}_{-\text{a}}\text{f(x)}\text{dx}=\begin{cases}2\int\limits^{\text{a}}_{0}\text{f(x)}\text{dx},&\text{ if }\text{f}(-\text{x})=\text{f(x)}\\0,&\text{ if }\text{f}(-\text{x})=-\text{f}(\text{x})\end{cases}\end{bmatrix}$
$=4\int\limits^{\pi}_0\frac{\text{x}\sin\text{x}}{1+\cos^2\text{x}}\text{ dx}\ ...(\text{i})$
Then,
$\text{I}_2=4\int\limits^{\pi}_0\frac{(\pi-\text{x})\sin(\pi-\text{x})}{1+\cos^2(\pi-\text{x})}\text{ dx}$
$=4\int\limits^{\pi}_0\frac{(\pi-\text{x})\sin\text{x}}{1+\cos^2\text{x}}\text{ dx}\ ....(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}_2=4\int\limits^{\pi}_0\frac{\pi\sin\text{x}}{1+\cos^2\text{x}}\text{ dx}$
$\Rightarrow2\text{I}_2=4\int\limits^{\pi}_0\frac{\sin\text{x}}{1+\cos^2\text{x}}\text{ dx}$
Put $\cos\text{x}=\text{z}$
$\Rightarrow-\sin\text{x}\text{dx}=\text{dz}$
When $\text{x}\rightarrow0,\text{ z}\rightarrow1$
When $\text{x}\rightarrow\pi,\text{ z}\rightarrow-1$
$\therefore\ 2\text{I}_2=-4\pi\int\limits^{-1}_1\frac{\text{dz}}{1+\text{z}^2}$
$\Rightarrow2\text{I}_2=-4\pi\times\Big[\tan^{-1}\text{z}\Big]^{-1}_1$
$\Rightarrow2\text{I}_2=-4\pi\Big[\tan^{-1}(-1)-\tan^{-1}1\Big]$
$\Rightarrow2\text{I}_2=-4\pi\Big(-\frac{\pi}{4}-\frac{\pi}{4}\Big)=2\pi^2$
$\Rightarrow\text{I}_2=\pi^2$
$\therefore\ \text{I}=\text{I}_1+\text{I}_2$
$\Rightarrow\text{I}=0+\pi^2=\pi^2$

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