Q118 / 160 · AP EAMCET 2016 · Mathematics

$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(\frac{x+\frac{\pi}{4}}{2-\cos 2 x}\right) d x$ is equal to

  • A
    $\frac{8 \pi \sqrt{3}}{5}$
  • B
    $\frac{2 \pi \sqrt{3}}{9}$
  • C
    $\frac{4 \pi^2 \sqrt{3}}{9}$
  • Correct answer
    D
    $\frac{\pi^2}{6 \sqrt{3}}$

Solution

Given, $\mathrm{I}=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x+\frac{\pi}{4}}{2-\cos ^2 x} d x$ $\mathrm{I}=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x+\frac{\pi}{4}}{2-\cos ^2 x} d x+\frac{\pi}{4} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{2-\cos ^2 x} d x$ Let, $I_1=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x}{2-\cos ^2 x} d x$ $f(x)=\frac{x}{2-\cos ^2 x}$ $f(-x)=\frac{-x}{2-\cos 2(-x)}=\frac{-x}{2-\cos 2 x}=-f(x)$ $f(x)$ is an odd function. Thus, $I_1=0$ Let $I_1=\frac{\pi}{4} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{2-\cos 2 x} d x$ $g(x)=\frac{\pi}{4(2-\cos 2 x)}$ $g(-x)=\frac{\pi}{4(2-\cos 2(-x))}=\frac{\pi}{4(2-\cos 2 x)}=g(x)$ $\therefore \quad g(x)$ is an even function Thus, $\mathrm{I}_2=2 \times \frac{\pi}{4} \int_0^{\frac{\pi}{4}} \frac{d x}{2-\cos 2 x}$ Now, $\mathrm{I}=\mathrm{I}_2=\frac{\pi}{4} \int_0^{\frac{\pi}{4}} \frac{d x}{2-\frac{1-\tan ^2 x}{1+\tan ^2 x}}$ $\begin{aligned} & \mathrm{I}=\frac{\pi}{4} \int_0^{\frac{\pi}{4}} \frac{1+\tan ^2 x}{2+\tan ^2 x-1+\tan ^2 x} d x \\ & \mathrm{I}=\frac{\pi}{4} \int_0^{\frac{\pi}{4}} \frac{\sec ^2 x}{1+3 \tan ^2 x} d x \end{aligned}$ Put $\tan x=t \Rightarrow \sec ^2 x d x=d t$ Limits : $x=0 \quad \Rightarrow \mathrm{t}=0$ $\begin{aligned} & x=\frac{\pi}{4} \quad \Rightarrow \mathrm{t}=1 \\ & \therefore \quad \mathrm{I}=\frac{\pi}{4} \int_0^1 \frac{d t}{1+3 t^2} \\ & \mathrm{I}=\frac{\pi}{2} \times \frac{1}{\sqrt{3}}\left[\tan ^{-1} \sqrt{3} t\right]_0^1 \\ & \mathrm{I}=\frac{\pi}{2 \sqrt{3}}\left[\tan ^{-1}(\sqrt{3})-\tan ^{-1}(0)\right] \\ & \mathrm{I}=\frac{\pi}{2 \sqrt{3}}\left(\frac{\pi}{3}\right)=\frac{\pi^2}{6 \sqrt{3}} \end{aligned}$

Hence, the correct answer is option: 'D'.

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