Mathematics - Definite Integration Question with Solution | TestHub

Let,

$I={\int }_{\frac{-\pi }{4}}^{\frac{\pi }{4}}\frac{x+\frac{\pi }{4}}{2-\cos 2x}dx \dots \left(1\right)$

Now replacing $\mathrm{x}\to -\mathrm{x}$ we get,

$\Rightarrow I={\int }_{\frac{-\pi }{4}}^{\frac{\pi }{4}}\frac{-x+\frac{\pi }{4}}{2-\cos 2x}dx \dots \left(2\right)$

Now both the equation we get,

$\Rightarrow 2\mathrm{I}={\int }_{\frac{-\pi }{4}}^{\frac{\pi }{4}}\frac{\frac{\pi }{2}}{2-\cos 2x}dx$

$\Rightarrow I=\frac{\pi }{4}·2{\int }_0^{\frac{\pi }{4}}\frac{dx}{2-\cos 2x}dx \left\{\text{as} \cos 2x \text{is even function}\right\}$

$\Rightarrow \mathrm{I}=\frac{\pi }{4}·2{\int }_0^{\frac{\pi }{4}}\frac{\left(1+{\tan }^2\mathrm{x}\right)\mathrm{dx}}{2\left(1+{\tan }^2\mathrm{x}\right)-\left(1-{\tan }^2\mathrm{x}\right)}$

$\Rightarrow \mathrm{I}=\frac{\pi }{4}·2{\int }_0^{\frac{\pi }{4}}\frac{{\sec }^{2}x\mathrm{dx}}{3{\tan }^2\mathrm{x}+1}$

Now let $\tan x=t\Rightarrow {\sec }^{2}xdx=dt$

$\Rightarrow I=\frac{\pi }{2}{\int }_0^1\frac{dt}{3t^2+1}$

$\Rightarrow \mathrm{I}=\frac{\pi }{2\sqrt{3}}{\tan }^{-1}\sqrt{3}$

$\Rightarrow I=\frac{{\pi }^2}{6\sqrt{3}}$