Question

If $I=\frac{2}{\mathrm{\pi }}{\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}\frac{1}{\left(1+e^{\sin \left(x\right)}\right)\left(2-\cos \left(2x\right)\right)}dx$ then $27I^2$ equals

Answer 1

Step 1: Simplify the given expression

Given: $I=\frac{2}{\mathrm{\pi }}{\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}\frac{1}{\left(1+e^{\sin \left(x\right)}\right)\left(2-\cos \left(2x\right)\right)}dx$ …$\left[1\right]$

$$\begin{gathered} \Rightarrow I=\frac{2}{\mathrm{\pi }}{\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}\frac{1}{\left(1+e^{\sin \left({\displaystyle \frac{\mathrm{\pi }}{4}}-{\displaystyle \frac{\mathrm{\pi }}{4}}-x\right)}\right)\left(2-\cos \left(2\left({\displaystyle \frac{\mathrm{\pi }}{4}}-{\displaystyle \frac{\mathrm{\pi }}{4}}-x\right)\right)\right)}dx\left[\because {\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx\right] \\ \Rightarrow I=\frac{2}{\mathrm{\pi }}{\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}\frac{1}{\left(1+e^{\sin \left({\displaystyle -x}\right)}\right)\left(2-\cos \left(2\left({\displaystyle -x}\right)\right)\right)}dx \\ \Rightarrow I=\frac{2}{\mathrm{\pi }}{\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}\frac{1}{\left(1+e^{-\sin \left({\displaystyle x}\right)}\right)\left(2-\cos \left(2{\displaystyle x}\right)\right)}dx\left[\because \sin (-x)=\sin \left(x\right),\cos (-x)=\cos \left(x\right)\right] \\ \Rightarrow I=\frac{2}{\mathrm{\pi }}{\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}\frac{1}{\left(1+{\displaystyle \frac{1}{e^{\sin \left(x\right)}}}\right)\left(2-\cos \left(2{\displaystyle x}\right)\right)}dx \\ \Rightarrow I=\frac{2}{\mathrm{\pi }}{\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}\frac{1}{\left({\displaystyle \frac{1+e^{\sin \left(x\right)}}{e^{\sin \left(x\right)}}}\right)\left(2-\cos \left(2{\displaystyle x}\right)\right)}dx \\ \Rightarrow I=\frac{2}{\mathrm{\pi }}{\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}\frac{e^{\sin \left(x\right)}}{\left({\displaystyle 1+e^{\sin \left(x\right)}}\right)\left(2-\cos \left(2{\displaystyle x}\right)\right)}dx \end{gathered}$$

Step 2: Simplify further

add equation $\left[1\right]$ to both sides

$$\begin{gathered} \Rightarrow I+I=\frac{2}{\mathrm{\pi }}{\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}\frac{e^{\sin \left(x\right)}}{\left({\displaystyle 1+e^{\sin \left(x\right)}}\right)\left(2-\cos \left(2{\displaystyle x}\right)\right)}dx+\frac{2}{\mathrm{\pi }}{\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}\frac{1}{\left({\displaystyle 1+e^{\sin \left(x\right)}}\right)\left(2-\cos \left(2{\displaystyle x}\right)\right)}dx \\ \Rightarrow 2I=\frac{2}{\mathrm{\pi }}{\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}\frac{1+e^{\sin \left(x\right)}}{\left({\displaystyle 1+e^{\sin \left(x\right)}}\right)\left(2-\cos \left(2{\displaystyle x}\right)\right)}dx \\ \Rightarrow I=\frac{1}{\mathrm{\pi }}{\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}\frac{1}{\left(2-\cos \left(2{\displaystyle x}\right)\right)}dx \\ \Rightarrow I=\frac{1}{\mathrm{\pi }}{\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}\frac{1}{\left(1+2{\sin }^2\left({\displaystyle x}\right)\right)}dx\left[\because \cos 2x=1-2{\sin }^2\left(x\right)\right] \\ \Rightarrow I=\frac{1}{\mathrm{\pi }}{\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}\frac{{\displaystyle \frac{1}{{\cos }^2\left(x\right)}}}{\left({\displaystyle \frac{1}{{\cos }^2\left(x\right)}}+2{\displaystyle \frac{{\sin }^2\left({\displaystyle x}\right)}{{\cos }^2\left(x\right)}}\right)}dx \\ \Rightarrow I=\frac{1}{\mathrm{\pi }}{\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}\frac{{\displaystyle {\sec }^2\left(x\right)}}{\left({\displaystyle {\sec }^2\left(x\right)+2{\tan }^2\left(x\right)}\right)}dx \\ \Rightarrow I=\frac{1}{\mathrm{\pi }}{\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}\frac{{\displaystyle {\sec }^2\left(x\right)}}{\left({\displaystyle 1+{\tan }^2\left(x\right)+2{\tan }^2\left(x\right)}\right)}dx\mathbf{}\mathbf{[}\mathbf{\because }{\mathbf{sec}}^{\mathbf{2}}\left(x\right)\mathbf{-}{\mathbf{tan}}^{\mathbf{2}}\left(x\right)\mathbf{=}\mathbf{1}\mathbf{]} \\ \Rightarrow I=\frac{1}{\mathrm{\pi }}{\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}\frac{{\displaystyle {\sec }^2\left(x\right)}}{\left({\displaystyle 1+3{\tan }^2\left(x\right)}\right)}dx \end{gathered}$$

Step 3: Simplify further using substitution

Let $\tan x=t$

$\Rightarrow se{c}^2\left(x\right)dx=dt$

$$\begin{gathered} \Rightarrow I=\frac{1}{\mathrm{\pi }}{\int }_{-1}^1\frac{{\displaystyle 1}}{\left({\displaystyle 1+3t^2}\right)}dt \\ \Rightarrow I=\frac{1}{3\mathrm{\pi }}{\int }_{-1}^1\frac{{\displaystyle 1}}{\left({\displaystyle \frac{{\displaystyle 1}}{3}+t^2}\right)}dt \\ \Rightarrow I=\frac{1}{3\mathrm{\pi }}{\int }_{-1}^1\frac{{\displaystyle 1}}{\left({\displaystyle {\left(\frac{{\displaystyle 1}}{\sqrt{3}}\right)}^2+{\left(t\right)}^2}\right)}dt \\ \Rightarrow I=\frac{1}{3\mathrm{\pi }}\sqrt{3}{\left|{\tan }^{-1}\left(\sqrt{3}t\right)\right|}_{-1}^1\left[\because \int \frac{1}{a^2+x^2}\mathrm{dx}=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)\right] \\ \Rightarrow I=\frac{1}{3\mathrm{\pi }}\sqrt{3}\left|{\tan }^{-1}\left(\sqrt{3}\left(1\right)\right)-{\tan }^{-1}\left(\sqrt{3}(-1)\right)\right| \\ \Rightarrow I=\frac{1}{3\mathrm{\pi }}\sqrt{3}\left|{\tan }^{-1}\left(\sqrt{3}\right)+{\tan }^{-1}\left(\sqrt{3}\right)\right| \\ \Rightarrow I=\frac{1}{3\mathrm{\pi }}2\sqrt{3}\times \frac{\mathrm{\pi }}{3} \\ \Rightarrow I=\frac{2}{3\sqrt{3}} \end{gathered}$$

Step 4: Solve for the required value

$$\begin{gathered} 27I^2=27{\left(\frac{2}{3\sqrt{3}}\right)}^2 \\ \Rightarrow 27I^2=27\left(\frac{4}{27}\right) \\ \Rightarrow 27I^2=4 \end{gathered}$$

Hence, the value of $27I^2$ is $4$.