Question

Given that the value of $I=\dfrac{2}{\pi }\int\limits_{\dfrac{-\pi }{4}}^{\dfrac{\pi }{4}}{\dfrac{dx}{\left( 1+{{e}^{\sin x}} \right)\left( 2-\cos 2x \right)}}dx$ then find the value of $27{{I}^{2}}$

Answer 1

To solve this question, we will first of all calculate the value of I using several integral formulas and properties. Firstly we will use
$\int\limits_{b}^{a}{f\left( x \right)dx=\int\limits_{a}^{b}{f\left( a+b-x \right)}}dx$
Then, we will obtain 2 values of I and add them up. We have now obtained value of 2I then we will use several trigonometric identity as stated $\cos 2\theta =2{{\cos }^{2}}\theta -1,{{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta \text{ and }\cos \theta =\dfrac{1}{\sec \theta }$. After applying all this, we will simplify value of 2I and then apply integration formula $\int\limits_{b}^{a}{\dfrac{1}{{{x}^{2}}+{{a}^{2}}}dx=\dfrac{1}{a}{{\tan }^{-1}}}\dfrac{x}{a}$ to get results.
$I=\dfrac{2}{\pi }\int\limits_{\dfrac{-\pi }{4}}^{\dfrac{\pi }{4}}{\dfrac{dx}{\left( 1+{{e}^{\sin x}} \right)\left( 2-\cos 2x \right)}}dx\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)}$

Complete step-by-step answer :
We have a property of definite integral given as below:
$\int\limits_{b}^{a}{f\left( x \right)dx=\int\limits_{a}^{b}{f\left( a+b-x \right)}}dx$
So, we can replace x by a+b-x then, we have $a=\dfrac{-\pi }{4}\text{ and b}=\dfrac{\pi }{4}$
Then, $a+b-x=\dfrac{-\pi }{4}+\dfrac{\pi }{4}-x=-x$
Using this value of a+b-x= -x and above stated property of definite integral we have;
$I=\dfrac{2}{\pi }\int\limits_{\dfrac{-\pi }{4}}^{\dfrac{\pi }{4}}{\dfrac{dx}{\left( 1+{{e}^{-\sin x}} \right)\left( 2-\cos 2x \right)}}$
Now, $\cos \left( -\theta \right)=\cos \theta$ Using this in above we get:
$I=\dfrac{2}{\pi }\int\limits_{\dfrac{-\pi }{4}}^{\dfrac{\pi }{4}}{\dfrac{dx}{\left( 1+{{e}^{-\sin x}} \right)\left( 2-\cos 2x \right)}}$
Writing ${{e}^{-\sin x}}=\dfrac{1}{{{e}^{\sin x}}}$ we get:

& I=\dfrac{2}{\pi }\int\limits_{\dfrac{-\pi }{4}}^{\dfrac{\pi }{4}}{\dfrac{dx}{\left( 1+\dfrac{1}{{{e}^{\sin x}}} \right)\left( 2-\cos 2x \right)}} \\\ & I=\dfrac{2}{\pi }\int\limits_{\dfrac{-\pi }{4}}^{\dfrac{\pi }{4}}{\dfrac{dx}{\left( \dfrac{{{e}^{\sin x}}+1}{{{e}^{\sin x}}} \right)\left( 2-\cos 2x \right)}} \\\ & I=\dfrac{2}{\pi }\int\limits_{\dfrac{-\pi }{4}}^{\dfrac{\pi }{4}}{\dfrac{{{e}^{\sin x}}}{\left( 1+{{e}^{\sin x}} \right)\left( 2-\cos 2x \right)}}dx\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (ii)} \\\ \end{aligned}$$ Now adding equation (i) and (ii) we get: $$2I=\dfrac{2}{\pi }\int\limits_{\dfrac{-\pi }{4}}^{\dfrac{\pi }{4}}{\dfrac{1+{{e}^{\sin x}}}{\left( 1+{{e}^{\sin x}} \right)\left( 2-\cos 2x \right)}}dx$$ Cancelling common term $\left( 1+{{e}^{\sin x}} \right)$ we get: $$2I=\dfrac{2}{\pi }\int\limits_{\dfrac{-\pi }{4}}^{\dfrac{\pi }{4}}{\dfrac{1}{\left( 2-\cos 2x \right)}}dx$$ Now, we have a trigonometric identity of $\cos \theta $ as $\cos 2\theta =-1+2{{\cos }^{2}}\theta $ Using this in above we get: $$\begin{aligned} & 2I=\dfrac{2}{\pi }\int\limits_{\dfrac{-\pi }{4}}^{\dfrac{\pi }{4}}{\dfrac{1}{\left( 2-\left( -1+2{{\cos }^{2}}x \right) \right)}}dx \\\ & 2I=\dfrac{2}{\pi }\int\limits_{\dfrac{-\pi }{4}}^{\dfrac{\pi }{4}}{\dfrac{1}{\left( 3-2{{\cos }^{2}}x \right)}}dx \\\ \end{aligned}$$ Now, dividing both numerator and denominator by ${{\cos }^{2}}x$ and using $\dfrac{1}{{{\cos }^{2}}x}=se{{x}^{2}}x$ in numerator we get: $$\begin{aligned} & 2I=\dfrac{2}{\pi }\int\limits_{\dfrac{-\pi }{4}}^{\dfrac{\pi }{4}}{\dfrac{{{\sec }^{2}}x}{\dfrac{3-2{{\cos }^{2}}x}{{{\cos }^{2}}x}}}dx \\\ & 2I=\dfrac{2}{\pi }\int\limits_{\dfrac{-\pi }{4}}^{\dfrac{\pi }{4}}{\dfrac{{{\sec }^{2}}x}{3{{\sec }^{2}}x-2}}dx \\\ \end{aligned}$$ Now, we have a trigonometric identity relating ${{\sec }^{2}}\theta \text{ and ta}{{\text{n}}^{\text{2}}}\theta $ as ${{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta $ in above we get: $$\begin{aligned} & 2I=\dfrac{2}{\pi }\int\limits_{\dfrac{-\pi }{4}}^{\dfrac{\pi }{4}}{\dfrac{{{\sec }^{2}}x}{3\left( 1+{{\tan }^{2}}\theta \right)-2}}dx \\\ & 2I=\dfrac{2}{\pi }\int\limits_{\dfrac{-\pi }{4}}^{\dfrac{\pi }{4}}{\dfrac{{{\sec }^{2}}x}{1+3{{\tan }^{2}}x}}dx \\\ \end{aligned}$$ Let tan x = t Differentiating with respect to x both sides we get: $${{\sec }^{2}}x=\dfrac{dt}{dx}\Rightarrow se{{x}^{2}}xdx=dt$$ When $x=\dfrac{-\pi }{4}$ then $\tan x=\tan \left( \dfrac{-\pi }{4} \right)=-1$ and when $x=\dfrac{\pi }{4}$ then $\tan x=\tan \left( \dfrac{\pi }{4} \right)=1$v Using this all substitution in above we get: $$2I=\dfrac{2}{\pi }\int\limits_{-1}^{+1}{\dfrac{dt}{1+3{{t}^{2}}}}$$ Taking $\dfrac{1}{3}$ common: $$\begin{aligned} & 2I=\dfrac{2}{\pi }\int\limits_{-1}^{+1}{\left( \dfrac{1}{3} \right)\dfrac{dt}{\left( {{\left( \dfrac{1}{\sqrt{3}} \right)}^{2}}+{{t}^{2}} \right)}} \\\ & 2I=\dfrac{2}{\pi }\int\limits_{-1}^{+1}{\dfrac{dt}{{{\left( \dfrac{1}{\sqrt{3}} \right)}^{2}}+{{t}^{2}}}} \\\ \end{aligned}$$ We have a integral property as: $$\int\limits_{c}^{b}{\dfrac{dt}{{{x}^{2}}+{{a}^{2}}}=\dfrac{1}{a}{{\tan }^{-1}}}\left. \dfrac{x}{a} \right|_{c}^{b}$$ Using this in above we get: $$\begin{aligned} & 2I=\dfrac{2}{3\pi }\left( \dfrac{1}{\dfrac{1}{\sqrt{3}}}\left( {{\tan }^{-1}}\sqrt{3}t \right) \right)_{-1}^{+1} \\\ & 2I=\dfrac{2}{3\pi }\times \sqrt{3}\left( {{\tan }^{-1}}\sqrt{3}t \right)_{-1}^{+1} \\\ & 2I=\dfrac{2}{3\pi }\sqrt{3}\left\\{ {{\tan }^{-1}}\sqrt{3}-{{\tan }^{-1}}\sqrt{3}\left( -1 \right) \right\\} \\\ & 2I=\dfrac{2}{3\pi }\sqrt{3}\left\\{ {{\tan }^{-1}}\sqrt{3}-{{\tan }^{-1}}\left( -\sqrt{3} \right) \right\\} \\\ \end{aligned}$$ Using $\tan \left( -\theta \right)=-\tan \theta $ we get: $$\begin{aligned} & 2I=\dfrac{2}{3\pi }\sqrt{3}\left\\{ {{\tan }^{-1}}\sqrt{3}+{{\tan }^{-1}}\sqrt{3} \right\\} \\\ & 2I=\dfrac{2}{3\pi }\sqrt{3}\text{ }2{{\tan }^{-1}}\sqrt{3} \\\ \end{aligned}$$ Now, we have $\tan \dfrac{\pi }{3}=\sqrt{3}$ applying ${{\tan }^{-1}}$ both sides we get: $$\begin{aligned} & \dfrac{\pi }{3}={{\tan }^{-1}}\left( \sqrt{3} \right) \\\ & 2I=\dfrac{2}{3\pi }\sqrt{3}\dfrac{\pi }{3}2 \\\ & I=\dfrac{2\sqrt{3}}{9} \\\ \end{aligned}$$ Now, we have obtained $I=\dfrac{2\sqrt{3}}{9}$ Then, value of $$\begin{aligned} & 27{{I}^{2}}=27{{\left( \dfrac{2\sqrt{3}}{9} \right)}^{2}} \\\ & \Rightarrow \dfrac{27\times 4\times 3}{9\times 9} \\\ & \Rightarrow \dfrac{3\times 3\times 4}{9}=4 \\\ \end{aligned}$$ Therefore, the value of $$27{{I}^{2}}=4$$ **Note** : Another way to solve after step $$2I=\dfrac{2}{\pi }\int\limits_{-1}^{+1}{\dfrac{dt}{1+3{{t}^{2}}}}$$ Let us assume ${{s}^{2}}=3{{t}^{2}}$ then differentiating with respect to t we get: $$\begin{aligned} & 2ds=3\times 2dt \\\ & \Rightarrow \dfrac{2ds}{6}=dt\Rightarrow \dfrac{ds}{3}=dt \\\ \end{aligned}$$ Also as ${{s}^{2}}=3{{t}^{2}}$ taking square root both sides $s=\pm \sqrt{3}$ When t = -1 then $s=\pm \sqrt{3}\left( -1 \right)=\pm -\sqrt{3}=\mp \sqrt{3}$ When t = +1 then $s=\pm \sqrt{3}$ So, we have $$t=-1\Rightarrow s=-\sqrt{3}\text{ and }t=1\Rightarrow s+\sqrt{3}$$ as integral units are always from lower to higher $$2I=\dfrac{2}{3\pi }\int\limits_{-\sqrt{3}}^{+\sqrt{3}}{\dfrac{1}{3}\dfrac{ds}{1+{{s}^{2}}}}$$ Now, we can use formula $$\int\limits_{c}^{b}{\dfrac{dt}{{{x}^{2}}+{{a}^{2}}}={{\tan }^{-1}}}\left. \dfrac{x}{a} \right|_{c}^{b}$$ result would anyway be the same.