Question

Evaluate the following integrals:

${\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\frac{2x\left(1+\sin x\right)}{1+{\cos }^{2}x}dx$

Answer 1

Let I = ${\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\frac{2x\left(1+\sin x\right)}{1+{\cos }^{2}x}dx$
Then,
$$\begin{gathered} I={\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\frac{2x}{1+{\cos }^{2}x}dx+{\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\frac{2x\sin x}{1+{\cos }^{2}x}dx \\ =I_1+I_2 \end{gathered}$$

Consider $f\left(x\right)=\frac{2x}{1+{\cos }^{2}x}$.
Now,
$f\left(-x\right)=\frac{2\left(-x\right)}{1+{\cos }^2\left(\mathrm{\pi }-x\right)}=-\frac{2x}{1+{\left(-\cos x\right)}^2}=-\frac{2x}{1+{\cos }^{2}x}=-f\left(x\right)$
$\therefore I_1={\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\frac{2x}{1+{\cos }^{2}x}dx=0\left[{\int }_{-a}^{a}f\left(x\right)dx=\left\{\begin{array}{ll}2{\int }_0^{a}f\left(x\right)dx,& \mathrm{if}f\left(-x\right)=f\left(x\right)\\ 0,& \mathrm{if}f\left(-x\right)=-f\left(x\right)\end{array}\right.\right]$

Again, consider $g\left(x\right)=\frac{2x\sin x}{1+{\cos }^{2}x}$.

$g\left(-x\right)=\frac{2\left(-x\right)\sin \left(-x\right)}{1+{\cos }^2\left(-x\right)}=\frac{2x\sin x}{1+{\cos }^{2}x}=g\left(x\right)\left[\sin \left(-x\right)=-\sin x\mathrm{and}\cos \left(-x\right)=\cos x\right]$
$$\begin{gathered} \therefore I_2={\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\frac{2x\sin x}{1+{\cos }^{2}x}dx \\ =2\times 2{\int }_0^{\mathrm{\pi }}\frac{x\sin x}{1+{\cos }^{2}x}dx\left[{\int }_{-a}^{a}f\left(x\right)dx=\left\{\begin{array}{ll}2{\int }_0^{a}f\left(x\right)dx,& \mathrm{if}f\left(-x\right)=f\left(x\right)\\ 0,& \mathrm{if}f\left(-x\right)=-f\left(x\right)\end{array}\right.\right] \\ =4{\int }_0^{\mathrm{\pi }}\frac{x\sin x}{1+{\cos }^{2}x}dx.....\left(1\right) \end{gathered}$$
Then,
$I_2=4{\int }_0^{\mathrm{\pi }}\frac{\left(\mathrm{\pi }-x\right)\sin \left(\mathrm{\pi }-x\right)}{1+{\cos }^2\left(\mathrm{\pi }-x\right)}dx=4{\int }_0^{\mathrm{\pi }}\frac{\left(\mathrm{\pi }-x\right)\sin x}{1+{\cos }^{2}x}dx.....\left(2\right)\left[{\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f\left(a-x\right)dx\right]$

Adding (1) and (2), we get

$$\begin{gathered} 2I_2=4{\int }_0^{\mathrm{\pi }}\frac{\mathrm{\pi }\sin x}{1+{\cos }^{2}x}dx \\ \Rightarrow 2I_2=4\mathrm{\pi }{\int }_0^{\mathrm{\pi }}\frac{\sin x}{1+{\cos }^{2}x}dx \end{gathered}$$

Put cosx = z

$\Rightarrow -\sin xdx=dz$

When $x\to 0,z\to 1$

When $x\to \mathrm{\pi },z\to -1$

$$\begin{gathered} \therefore 2I_2=-4\mathrm{\pi }{\int }_1^{-1}\frac{dz}{1+z^2} \\ \Rightarrow 2I_2=-4\mathrm{\pi }\times {\overline{){\tan }^{-1}z}}_1^{-1} \\ \Rightarrow 2I_2=-4\mathrm{\pi }\left[{\tan }^{-1}\left(-1\right)-{\tan }^{-1}1\right] \\ \Rightarrow 2I_2=-4\mathrm{\pi }\left(-\frac{\mathrm{\pi }}{4}-\frac{\mathrm{\pi }}{4}\right)=2{\mathrm{\pi }}^2 \\ \Rightarrow I_2={\mathrm{\pi }}^2 \end{gathered}$$

$\therefore I=I_1+I_2=0+{\mathrm{\pi }}^2={\mathrm{\pi }}^2$