Mathematics - Definite Integration Question with Solution | TestHub

$I=\frac{2}{\pi }\underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}}{\int }}\frac{1}{\left(1+e^{\mathrm{s}\mathrm{i}\mathrm{n}x}\right)\left(2-\mathrm{c}\mathrm{o}\mathrm{s}2x\right)}\cdot dx$
Using$\underset{-a}{\overset{a}{\int }}f\left(x\right)\cdot dx=\underset{0}{\overset{a}{\int }}\left(f\left(x\right)+f\left(-x\right)\right)\cdot dx$
$I=\frac{2}{\pi }\underset{0}{\overset{\frac{\pi }{4}}{\int }}\left(\frac{1}{\left(1+e^{\mathrm{s}\mathrm{i}\mathrm{n}x}\right)\left(2-\mathrm{c}\mathrm{o}\mathrm{s}2x\right)}+\frac{1}{\left(1+e^{-\mathrm{s}\mathrm{i}\mathrm{n}x}\right)\left(2-\mathrm{c}\mathrm{o}\mathrm{s}2x\right)}\right)\cdot dx$
$I=\frac{2}{\pi }\underset{0}{\overset{\frac{\pi }{4}}{\int }}\left(\frac{1}{\left(1+e^{\mathrm{s}\mathrm{i}\mathrm{n}x}\right)\left(2-\mathrm{c}\mathrm{o}\mathrm{s}2x\right)}+\frac{e^{\mathrm{s}\mathrm{i}\mathrm{n}x}}{\left(e^{\mathrm{s}\mathrm{i}\mathrm{n}x}+1\right)\left(2-\mathrm{c}\mathrm{o}\mathrm{s}2x\right)}\right)\cdot dx$
$I=\frac{2}{\pi }\underset{0}{\overset{\frac{\pi }{4}}{\int }}\frac{dx}{2-\mathrm{c}\mathrm{o}\mathrm{s}2x}$
$I=\frac{2}{\pi }\underset{0}{\overset{\frac{\pi }{4}}{\int }}\frac{dx}{2-\frac{1-{\mathrm{t}\mathrm{a}\mathrm{n}}^{2}x}{1+{\mathrm{t}\mathrm{a}\mathrm{n}}^{2}x}}=\frac{2}{\pi }\underset{0}{\overset{\frac{\pi }{4}}{\int }}\frac{\left(1+{\mathrm{t}\mathrm{a}\mathrm{n}}^{2}x\right)dx}{1+3{\mathrm{t}\mathrm{a}\mathrm{n}}^{2}x}$
$I=\frac{2}{\pi }\underset{0}{\overset{\frac{\pi }{4}}{\int }}\frac{{\mathrm{s}\mathrm{e}\mathrm{c}}^{2}xdx}{1+3ta{n}^{2}x}$
Put$\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{ }x=t$
${\sec }^{2}xdx=dt$
and when$x=0\Rightarrow t=0$
$x=\frac{\pi }{4}\Rightarrow t=1$
$I=\frac{2}{\pi }\underset{0}{\overset{1}{\int }}\frac{dt}{1+3t^2}$
$I=\frac{2}{\pi }\underset{0}{\overset{1}{\int }}\frac{dt}{1+{\left(\sqrt{3}t\right)}^2}$
$I=\frac{2}{\pi }\frac{1}{\sqrt{3}}{\left[\mathrm{t}\mathrm{a}{\mathrm{n}}^{-1}\sqrt{3}t\right]}_0^1 \mathrm{ }\left(\because \underset{0}{\overset{1}{\int }}\frac{dx}{a^2+x^2}=\frac{1}{a}{\tan }^{-1}\frac{x}{a}+C\right)$
$I=\frac{2}{\sqrt{3}\pi }\left[\mathrm{t}\mathrm{a}{\mathrm{n}}^{-1}\left(\sqrt{3}\right)-\mathrm{t}\mathrm{a}{\mathrm{n}}^{-1}\left(0\right)\right]$
$I=\frac{2}{\sqrt{3}\pi }\times \frac{\pi }{3}$
$I=\frac{2}{3\sqrt{3}}$
$27I^2=27\times \frac{4}{27}=4$