Question

The integral $\underset{\frac{\pi }{6}}{\overset{\frac{\pi }{4}}{\int }}\frac{dx}{\sin 2x({\tan }^{5}x+co{t}^{5}x)}$ 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. $\frac{1}{20}{\tan }^{-1}(\frac{1}{9\sqrt{3}})$

Answer 1

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

$I=\underset{\frac{\pi }{6}}{\overset{\frac{\pi }{4}}{\int }}\frac{dx}{\sin 2x({\tan }^{5}x+co{t}^{5}x)}$

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

$I=\underset{\frac{\pi }{6}}{\overset{\frac{\pi }{4}}{\int }}\frac{{\sec }^{2}xdx}{2\tan x\left(\frac{{\tan }^{10}x+1}{{\tan }^{5}x}\right)}$
Let $\tan x=t\Rightarrow {\sec }^{2}xdx=dt$

$I=\underset{\frac{1}{\sqrt{3}}}{\overset{1}{\int }}\frac{t^{5}dt}{2t(t^{10}+1)}$
​​​​​​​​​​​​​​$I=\frac{1}{2}\underset{\frac{1}{\sqrt{3}}}{\overset{1}{\int }}\frac{t^{4}dt}{(t^{10}+1)}$
Let $t^5=k\Rightarrow 5t^{4}dt=dk$
​​​​​​​​​​​​​​$I=\frac{1}{10}\underset{(\frac{1}{\sqrt{3}}{)}^5}{\overset{1}{\int }}\frac{dk}{k^2+1}dk$
$I=\frac{1}{10}{\left[{\tan }^{-1}k\right]}_{{\left(\frac{1}{\sqrt{3}}\right)}^5}^1$

$I=\frac{1}{10}(\frac{\pi }{4}-{\tan }^{-1}\frac{1}{9\sqrt{3}})$