Question

The integral $\underset{\pi /6}{\overset{\pi /4}{\int }}\frac{dx}{\sin 2x({\tan }^{5}x+{\cot }^{5}x)}$ 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. $\frac{\pi }{40}$
• D. $\frac{1}{20}\left[{\tan }^{-1}\left(\frac{1}{9\sqrt{3}}\right)\right]$

Answer 1

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

$I=\underset{\pi /6}{\overset{\pi /4}{\int }}\frac{dx}{\sin 2x({\tan }^{5}x+{\cot }^{5}x)}$
$\Rightarrow I=\underset{\pi /6}{\overset{\pi /4}{\int }}\frac{(1+{\tan }^{2}x)dx}{2\tan x({\tan }^{5}x+\frac{1}{{\tan }^{5}x})}$

Put $\tan x=t\Rightarrow \{\begin{array}{cc}t=\tan \frac{\pi }{6}=\frac{1}{\sqrt{3}}& \\ t=\tan \frac{\pi }{4}=1& \end{array}$
$\Rightarrow {\sec }^{2}xdx=dt$
$\Rightarrow (1+{\tan }^{2}x)dx=dt$

$\therefore I=\underset{1/\sqrt{3}}{\overset{1}{\int }}\frac{1}{2t(t^5+\frac{1}{t^5})}dt$
$\Rightarrow I=\underset{1/\sqrt{3}}{\overset{1}{\int }}\frac{t^4}{2(t^{10}+1)}dt$

Put $t^5=u\Rightarrow 5t^{4}dt=du$
$\Rightarrow I=\frac{1}{10}\underset{3^{-5/2}}{\overset{1}{\int }}\frac{1}{u^2+1}du$

$\Rightarrow I={\left[\frac{1}{10}{\tan }^{-1}u\right]}_{3^{-5/2}}^1$

$=\frac{1}{10}[\frac{\pi }{4}-{\tan }^{-1}{3}^{-5/2}]$

$=\frac{1}{10}[\frac{\pi }{4}-{\tan }^{-1}\left(\frac{1}{9\sqrt{3}}\right)]$