Question

Evaluate ${\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{x+\frac{\pi }{4}}{2-cos2x}dx.$

Answer 1

${\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{x+\frac{\pi }{4}}{2-cos2x}dx={\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{x}{2-cos2x}dx+\frac{\pi }{4}{\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{1}{2-cos2x}dx.$
Consider $f\left(x\right)=\frac{x}{2-cos2x}\Rightarrow f(-x)=\frac{-x}{2-cos2(-x)}=-\frac{x}{2-cos2x}=-f\left(x\right)$
And, $g\left(x\right)=\frac{1}{2-cos2x}\Rightarrow g(-x)=\frac{1}{2-cos2(-x)}=\frac{1}{2-cos2x}=g\left(x\right).$
Clearly f(x) and g(x) are respectively odd and even functions.
By using ${\int }_{-a}^{a}f\left(x\right)dx=\{\begin{array}{cc}0,iff\left(x\right)isodd& \\ 2{\int }_0^{a}f\left(x\right)dx,iff\left(x\right)iseven& \end{array}$, we get :
$I=0+\frac{\pi }{4}\times {\int }_0^{\frac{\pi }{4}}\frac{1}{2-cos2x}dx=\frac{\pi }{2}{\int }_0^{\frac{\pi }{4}}\frac{1}{2-\frac{1-ta{n}^{2}x}{1+ta{n}^{2}x}}dx=\frac{\pi }{2}{\int }_0^{\frac{\pi }{4}}\frac{se{c}^{2}x}{3ta{n}^{2}x+1}$
Put $tanx=t\Rightarrow se{c}^{2}xdx=dt.$
When $x=0\Rightarrow t=0,whenx=\frac{\pi }{4}\Rightarrow t=1.$
So, $I=\frac{\pi }{2}{\int }_0^1\frac{1}{3t^2+1}dt=\frac{\pi }{2}\times \frac{1}{\sqrt{3}}\left[ta{n}^{-1}\right(\sqrt{3}t){]}_0^1=\frac{\pi }{2\sqrt{3}}[ta{n}^{-1}\sqrt{3}-0]=\frac{{\pi }^2}{6\sqrt{3}}.$