Question

The value of $\underset{\frac{-\pi }{3}}{\overset{\frac{\pi }{3}}{\int }}\frac{\pi +4x^3}{2-\cos (\left|x\right|+\frac{\pi }{3})}dx$ is

• A. $-\frac{2\pi }{3}{\tan }^{-1}\left(\frac{1}{3}\right)$
• B. $\frac{4\pi }{\sqrt{3}}{\tan }^{-1}\left(\frac{1}{2}\right)$
• C. $\frac{\pi }{3}{\tan }^{-1}\left(\frac{1}{4}\right)$
• D. $\frac{2\pi }{3}{\tan }^{-1}\left(\frac{1}{5}\right)$

Answer 1

The correct option is B $\frac{4\pi }{\sqrt{3}}{\tan }^{-1}\left(\frac{1}{2}\right)$

$I=\underset{\frac{-\pi }{3}}{\overset{\frac{\pi }{3}}{\int }}\frac{\pi +4x^3}{2-\cos (\left|x\right|+\frac{\pi }{3})}\Rightarrow I=\underset{\frac{-\pi }{3}}{\overset{\frac{\pi }{3}}{\int }}\frac{\pi }{2-\cos (\left|x\right|+\frac{\pi }{3})}+\underset{\frac{-\pi }{3}}{\overset{\frac{\pi }{3}}{\int }}\frac{4x^3}{2-\cos (\left|x\right|+\frac{\pi }{3})}\Rightarrow I=2\underset{0}{\overset{\frac{\pi }{3}}{\int }}\frac{\pi }{2-\cos (\left|x\right|+\frac{\pi }{3})}dx+0[\because \underset{-a}{\overset{a}{\int }}f\left(x\right)dx=0,\text{when}f\left(x\right)\text{is odd}]\Rightarrow I=2\pi \underset{0}{\overset{\frac{\pi }{3}}{\int }}\frac{dx}{2-\cos (x+\frac{\pi }{3})}$

Putting $x+\frac{\pi }{3}=t\Rightarrow dx=dt$
$\Rightarrow I=2\pi \underset{\frac{\pi }{3}}{\overset{\frac{2\pi }{3}}{\int }}\frac{dt}{2-\cos t}\Rightarrow I=2\pi \underset{\frac{\pi }{3}}{\overset{\frac{2\pi }{3}}{\int }}\frac{dt}{2-\left(\frac{1-{\tan }^2\frac{t}{2}}{1+{\tan }^2\frac{t}{2}}\right)}\Rightarrow I=2\pi \underset{\frac{\pi }{3}}{\overset{\frac{2\pi }{3}}{\int }}\frac{{\sec }^2\frac{t}{2}dt}{1+3{\tan }^2\frac{t}{2}}$

Taking $\tan \frac{t}{2}=u\Rightarrow {\sec }^2\frac{t}{2}dt=2du$
$\Rightarrow I=2\pi \underset{\frac{1}{\sqrt{3}}}{\overset{\sqrt{3}}{\int }}\frac{2du}{3u^2+1}\Rightarrow I=2\pi \times \frac{2}{3}\underset{\frac{1}{\sqrt{3}}}{\overset{\sqrt{3}}{\int }}\frac{du}{u^2+{\left(\frac{1}{\sqrt{3}}\right)}^2}\Rightarrow I=\frac{4\pi }{3}\times \sqrt{3}{\left[{\tan }^{-1}\left(\frac{u}{\frac{1}{\sqrt{3}}}\right)\right]}_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}\Rightarrow I=\frac{4\pi }{\sqrt{3}}\left[{\tan }^{-1}\right(3)-{\tan }^{-1}(1\left)\right]\Rightarrow I=\frac{4\pi }{\sqrt{3}}{\tan }^{-1}\left(\frac{2}{1+3}\right)\therefore I=\frac{4\pi }{\sqrt{3}}{\tan }^{-1}\left(\frac{1}{2}\right)$