Question

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

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

Answer 1

The correct option is A $\frac{1}{10}(\frac{\pi }{4}-{\tan }^{-1}\left(\frac{1}{9\sqrt{3}}\right))$

$l={\int }_{\frac{\pi }{6}}^{\frac{\pi }{4}}\frac{{\sec }^{2}xdx}{2\tan x({\tan }^{5}x+{\cot }^{5}x)}$

$\tan x=t={\int }_{\frac{1}{\sqrt{3}}}^1\frac{dt}{2t(t^5+\frac{1}{t^5})}={\int }_{\frac{1}{\sqrt{3}}}^1\frac{t^{4}dt}{2(t^{10}+1)}$

$t^5=p\Rightarrow t^{4}dt=\frac{dp}{5}$

$={\int }_{3^{-\frac{5}{2}}}^1\frac{1}{2(p^2+1)}\frac{dp}{5}=\frac{1}{10}({\tan }^{-1}p{)}_{3^{-\frac{5}{2}}}^1=\frac{1}{10}(\frac{\pi }{4}-{\tan }^{-1}{3}^{\frac{-5}{2}})=\frac{1}{10}(\frac{\pi }{4}-{\tan }^{-1}\frac{1}{9\sqrt{3}})$