Question

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

• A. $\frac{{\mathrm{\pi }}^2}{4}$
• B. ${\mathrm{\pi }}^2$
• C. zero
• D. $\frac{\mathrm{\pi }}{2}$

Answer 1

The correct option is B

${\mathrm{\pi }}^2$

Explanation for the correct option:

Finding the value of the given integral:

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

Take $I_1={\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\left[\frac{2x}{\left(1+{\cos }^{2}x\right)}\right]dx$

$$\begin{gathered} \Rightarrow f\left(x\right)=\left[\frac{2x}{1+{\cos }^{2}x}\right] \\ \Rightarrow f\left(-x\right)=\left[\frac{2\left(-x\right)}{1+{\cos }^2\left(\mathrm{\pi }-\mathrm{x}\right)}\right] \\ =-\left[\frac{2\left(x\right)}{1+{\left(-\cos x\right)}^2}\right] \\ =-\left[\frac{2\left(x\right)}{1+{\cos }^{2}x}\right] \\ =-f\left(x\right) \\ \Rightarrow I_1={\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\frac{2x}{1+{\cos }^{2}x}dx \\ =0(\mathrm{since}\mathrm{f}(\mathrm{x})\mathrm{is}\mathrm{an}\mathrm{odd}\mathrm{function}) \end{gathered}$$

$$\begin{gathered} \mathrm{Again}\mathrm{consider} \\ \Rightarrow \mathrm{g}\left(\mathrm{x}\right)=\frac{2\mathrm{x}\mathrm{sinx}}{1+{\cos }^2\mathrm{x}} \\ \Rightarrow \mathrm{g}\left(-\mathrm{x}\right)=\frac{2\left(-\mathrm{x}\right)\sin \left(-\mathrm{x}\right)}{1+{\cos }^2\left(-\mathrm{x}\right)} \\ =\frac{2\left(\mathrm{x}\right)\sin \left(\mathrm{x}\right)}{1+{\cos }^2\left(\mathrm{x}\right)}\left[\mathrm{since}\sin \left(-\mathrm{x}\right)=-\mathrm{sinx},\cos \left(-\mathrm{x}\right)=\mathrm{cosx}\right] \\ =\mathrm{g}\left(\mathrm{x}\right) \\ \Rightarrow I_2={\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\frac{2x\sin x}{1+{\cos }^{2}x}dx \\ =2\times 2{\int }_0^{\mathrm{\pi }}\frac{x\sin x}{1+{\cos }^{2}x}dx \\ =4{\int }_0^{\mathrm{\pi }}\frac{x\sin x}{1+{\cos }^{2}x}dx.............\left(1\right)\left[\mathrm{since}\mathrm{g}\left(\mathrm{x}\right)\mathrm{is}\mathrm{an}\mathrm{even}\mathrm{function}\right] \\ \mathrm{Then}, \\ \Rightarrow I_2=4{\int }_0^{\mathrm{\pi }}\frac{\left(\mathrm{\pi }-\mathrm{x}\right)\sin \left(\mathrm{\pi }-\mathrm{x}\right)}{1+{\cos }^2\left(\mathrm{\pi }-\mathrm{x}\right)}dx \\ =4{\int }_0^{\mathrm{\pi }}\frac{\left(\mathrm{\pi }-\mathrm{x}\right)\sin \left(\mathrm{x}\right)}{1+{\cos }^2\left(\mathrm{x}\right)}dx..........\left(2\right)\left[{\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f\left(a-x\right)dx\right] \end{gathered}$$

Adding $\left(1\right)$and $\left(2\right)$, we get

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

Substitute $\cos x=z$

$-\sin xdx=dz$

when $x\to 0,z\to 1$ and

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

$$\begin{gathered} \Rightarrow 2I_2=-4\mathrm{\pi }{\int }_1^{-1}\frac{dz}{1+z^2} \\ \Rightarrow 2{\mathrm{I}}_2=-4\mathrm{\pi }{\left[{\tan }^{-1}\mathrm{z}\right]}_1^{-1} \\ \Rightarrow 2{\mathrm{I}}_2=-4\mathrm{\pi }\left[-\frac{\mathrm{\pi }}{4}-\frac{\mathrm{\pi }}{4}\right] \\ \Rightarrow 2{\mathrm{I}}_2=2{\mathrm{\pi }}^2 \\ \Rightarrow {\mathrm{I}}_2={\mathrm{\pi }}^2 \end{gathered}$$

$$\begin{gathered} \Rightarrow I=I_1+I_2 \\ =0+{\mathrm{\pi }}^2 \\ ={\mathrm{\pi }}^2 \end{gathered}$$

Hence, Option (B) is the correct answer