Question

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

• A. \(\frac{\pi}{40}\)
• B. \(\frac{1}{20}\tan^{-1}\left(\frac{1}{9\sqrt3}\right)\)
• C. \(\frac{1}{10}\left(\frac{\pi}{4} - \tan^{-1}\left(\frac{1}{9\sqrt3}\right)\right)\)
• D. \(\frac{1}{5}\left(\frac{\pi}{4} - \tan^{-1}\left(\frac{1}{3\sqrt3}\right)\right)\)

Hint:

Substitute t = tan x, then u = t^5; the integrand reduces to du/(10(1+u^2)).

Answer 1

The Correct Option is C

Let \(t = \tan x\). Then \(dx = \frac{dt}{1+t^2}\) and \(\sin2x = \frac{2t}{1+t^2}\).
The integrand becomes \(\frac{dt}{1+t^2}\cdot\frac{1+t^2}{2t}\cdot\frac{1}{t^5 + t^{-5}} = \frac{t^4\,dt}{2(1+t^{10})}\).
Now put \(u = t^5\), \(du = 5t^4\,dt\): the integral is \(\frac{1}{10}\int\frac{du}{1+u^2}\).
Limits: \(x = \frac\pi6\) gives \(t = \frac{1}{\sqrt3}\), \(u = \frac{1}{9\sqrt3}\); \(x = \frac\pi4\) gives \(u = 1\).
Value = \(\frac{1}{10}\left[\tan^{-1}u\right]_{1/(9\sqrt3)}^{1} = \frac{1}{10}\left(\frac\pi4 - \tan^{-1}\frac{1}{9\sqrt3}\right)\).
Answer: (3) \(\frac{1}{10}\left(\frac{\pi}{4} - \tan^{-1}\left(\frac{1}{9\sqrt3}\right)\right)\)