Question:
The integral \(\displaystyle\int_{\pi/12}^{\pi/4} \frac{8\cos 2x}{(\tan x + \cot x)^3}\,dx\) equals :
The integral \(\displaystyle\int_{\pi/12}^{\pi/4} \frac{8\cos 2x}{(\tan x + \cot x)^3}\,dx\) equals :
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Use \(\tan x + \cot x = \frac{2}{\sin 2x}\) to reduce the integrand to \(\cos 2x\sin^3 2x\), then substitute \(t = \sin 2x\).
Updated On: Nov 11, 2026
- \(\frac{15}{128}\)
- \(\frac{15}{64}\)
- \(\frac{13}{32}\)
- \(\frac{13}{256}\)
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The Correct Option is A
Solution and Explanation
\(\tan x + \cot x = \frac{\sin^2 x + \cos^2 x}{\sin x\cos x} = \frac{2}{\sin 2x}\).
The integrand becomes \(\frac{8\cos 2x\,\sin^3 2x}{8} = \cos 2x\,\sin^3 2x\).
Put \(t = \sin 2x\), \(dt = 2\cos 2x\,dx\). Limits: \(x = \frac{\pi}{12}\) gives \(t = \frac{1}{2}\); \(x = \frac{\pi}{4}\) gives \(t = 1\).
Integral \(= \frac{1}{2}\int_{1/2}^{1} t^3\,dt = \frac{1}{2}\cdot\frac{1}{4}\left(1 - \frac{1}{16}\right) = \frac{1}{8}\cdot\frac{15}{16} = \frac{15}{128}\).
Answer: (1) \(\frac{15}{128}\)
The integrand becomes \(\frac{8\cos 2x\,\sin^3 2x}{8} = \cos 2x\,\sin^3 2x\).
Put \(t = \sin 2x\), \(dt = 2\cos 2x\,dx\). Limits: \(x = \frac{\pi}{12}\) gives \(t = \frac{1}{2}\); \(x = \frac{\pi}{4}\) gives \(t = 1\).
Integral \(= \frac{1}{2}\int_{1/2}^{1} t^3\,dt = \frac{1}{2}\cdot\frac{1}{4}\left(1 - \frac{1}{16}\right) = \frac{1}{8}\cdot\frac{15}{16} = \frac{15}{128}\).
Answer: (1) \(\frac{15}{128}\)
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