Question

Evaluate the definite integral : ${\int }_{-\pi }^{\pi }\frac{2x(1+sinx)}{1+co{s}^{2}x}dx.$

Answer 1

Let I = ${\int }_{-\pi }^{\pi }\frac{2x(1+sinx)}{1+co{s}^{2}x}dx\to I={\int }_{-\pi }^{\pi }\frac{2x}{1+co{s}^{2}x}dx+{\int }_{-\pi }^{\pi }\frac{2xsinx}{1+co{s}^{2}x}dx\Rightarrow f(-x)=\frac{-2x}{1+co{s}^{2}x}=-f\left(x\right)$
and $g\left(x\right)=\frac{2xsinx}{1+co{s}^{2}x}\Rightarrow g(-x)=\frac{2(-x)sin(-x)}{1+co{s}^2(-x)}=\frac{2xsinx}{1+co{s}^{2}x}=g\left(x\right)$
Clearly, f(x) is an odd function whereas g(x) is an even function.
So, I = 0+2 ${\int }_0^{\pi }\frac{2xsinx}{1+co{s}^{2}x}dx\Rightarrow I=4{\int }_0^{\pi }\frac{xsinx}{1+co{s}^{2}x}dx$..........(i)
$\Rightarrow I=4{\int }_0^{\pi }\frac{(\pi -x)sin(\pi -x)}{1+co{s}^2(\pi -x)}dx\Rightarrow I=4{\int }_0^{\pi }\frac{(\pi -x)sinx}{1+co{s}^{2}x}dx$....(ii)
On adding (i) & (ii), we get : $2I=4{\int }_0^{\pi }\frac{xsinx}{1+co{s}^{2}x}dx+4{\int }_0^{\pi }\frac{(\pi -x)sinx}{1+co{s}^{2}x}dx\Rightarrow 2I=4\pi {\int }_0^{\pi }\frac{sinx}{1+co{s}^{2}x}dx\Rightarrow I=2\pi {\int }_0^{\pi }\frac{sinx}{1+co{s}^{2}x}dx\text{Consider}f\left(x\right)=\frac{sinx}{1+co{s}^{2}x}\Rightarrow f(\pi -x)=\frac{sin(\pi -x)}{1+co{s}^2(\pi -x)}=\frac{sinx}{1+co{s}^{2}x}=f\left(x\right)\text{By using}{\int }_0^{2a}f\left(x\right)dx=\{\begin{array}{c}2{\int }_0^{a}f\left(x\right)dx,iff(2a-x)=f\left(x\right)\\ 0,iff(2a-x)=-f\left(x\right)\end{array}\text{We have}I=2\pi \times 2{\int }_0^{\frac{\pi }{2}}\frac{sinx}{1+co{s}^{2}x}dx\text{Put cos x=t}\Rightarrow sinxdx=-dt\text{When}x=0\Rightarrow t=1\&\text{when}x=\frac{\pi }{2}\Rightarrow t=0\therefore I=-2\pi \times 2{\int }_1^0\frac{dt}{1+t^2}\Rightarrow I=4\pi {\int }_0^1\frac{dt}{1+t^2}\Rightarrow I=4\pi [ta{n}^{-1}t{]}_0^1\Rightarrow I=4\pi [ta{n}^{-1}1-ta{n}^{-1}0]=4\pi [\frac{\pi }{4}-0]={\pi }^2.$