The integral int_{frac{pi }{12}}^{frac{pi }{4 | vedclass

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Question

(A) We know that $	an x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} = \frac{1}{\sin x \cos x

Vedclass · Accepted Answer

$\frac{15}{128}$

Vedclass · Answer

(A) We know that $	an x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} = \frac{1}{\sin x \cos x} = \frac{2}{\sin 2x}$.Substituting this into the integral:$I = \int_{\frac{\pi}{12}}^{\frac{\pi}{4}} \frac{8 \cos 2x}{(\frac{2}{\sin 2x})^3} dx = \int_{\frac{\pi}{12}}^{\frac{\pi}{4}} \frac{8 \cos 2x \sin^3 2x}{8} dx = \int_{\frac{\pi}{12}}^{\frac{\pi}{4}} \cos 2x \sin^3 2x dx$.Let $u = \sin 2x$,then $du = 2 \cos 2x dx$,so $\cos 2x dx = \frac{du}{2}$.When $x = \frac{\pi}{12}$,$u = \sin(\frac{\pi}{6}) = \frac{1}{2}$.When $x = \frac{\pi}{4}$,$u = \sin(\frac{\pi}{2}) = 1$.$I = \int_{1/2}^{1} u^3 \frac{du}{2} = \frac{1}{2} [\frac{u^4}{4}]_{1/2}^{1} = \frac{1}{8} [1^4 - (\frac{1}{2})^4] = \frac{1}{8} [1 - \frac{1}{16}] = \frac{1}{8} 	imes \frac{15}{16} = \frac{15}{128}$.

The integral $\int_{\frac{\pi }{12}}^{\frac{\pi }{4}} \frac{8 \cos 2x}{(\tan x + \cot x)^3} dx$ equals

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