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Q. The integral $\int\limits^{\frac{\pi}{4}}_{\frac{\pi}{12}} \frac{8 \cos 2x}{\left(\tan x + \cot x\right)^{3}} dx $ equals :

JEE MainJEE Main 2017Integrals

Solution:

$\int\limits^{\frac{\pi}{4}}_{\frac{\pi}{12}}$$\frac{cos 2x}{\left(\frac{1}{sin 2x}\right)^{3}} = $$\int\limits^{\frac{\pi}{4}}_{\frac{\pi}{12}}$$ cos \,2x \times sin\, 2x.sin^{2}\, \left(2x\right) \,dx$
$= \frac{1}{4}\int sin 4x.\left(1-cos 4x\right)dx$
$= \frac{1}{4}$$\int\limits^{\frac{\pi}{4}}_{\frac{\pi}{12}}$ sin 4x - $\int\limits^{\frac{\pi}{4}}_{\frac{\pi}{12}}$ sin 8x