Question

The integral ${\int }_{\frac{\pi }{6}}^{\frac{\pi }{4}}\frac{dx}{\sin 2x\left({\tan }^{5}x+{\cot }^{5}x\right)}$ equals :-

• A. $\frac{1}{10}\left(\frac{\pi }{4}-{\tan }^{-1}\left(\frac{1}{9\sqrt{3}}\right)\right)$
• B. $\frac{1}{5}\left(\frac{\pi }{4}-{\tan }^{-1}\left(\frac{1}{3\sqrt{3}}\right)\right)$
• C. $\frac{\pi }{10}$
• D. $\frac{1}{20}{\tan }^{-1}\left(\frac{1}{9\sqrt{3}}\right)$

Answer 1

Answer: (A)

$I={\int }_{\pi /6}^{\pi /4}\frac{dx}{\sin 2x\left({\tan }^{5}x+{\cot }^{5}x\right)}$
$I=\frac{1}{2}{\int }_{\pi /6}^{\pi /4}\frac{{\tan }^{4}x{\sec }^{2}xdx}{\left(1+{\tan }^{10}x\right)}$ Put ${\tan }^{5}x=t$
$I=\frac{1}{10}{\int }_{{\left(\frac{1}{\sqrt{3}}\right)}^5}^1\frac{dt}{1+t^2}=\frac{1}{10}\left(\frac{\pi }{4}-{\tan }^{-1}\frac{1}{9\sqrt{3}}\right)$