int frac{dx}{4cos^3 2x - 3cos 2x} = | vedclass

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Question

(B) We know the trigonometric identity $\cos 3	heta = 4\cos^3 	heta - 3\cos 	heta$. Substituting $	heta = 2x$,we get $\cos 6x = 4\cos^3 2x - 3\cos

Vedclass · Accepted Answer

$\frac{1}{6}\log |\sec 6x + 	an 6x| + c$

Vedclass · Answer

(B) We know the trigonometric identity $\cos 3	heta = 4\cos^3 	heta - 3\cos 	heta$. Substituting $	heta = 2x$,we get $\cos 6x = 4\cos^3 2x - 3\cos 2x$. Therefore,the integral becomes: $\int \frac{dx}{4\cos^3 2x - 3\cos 2x} = \int \frac{dx}{\cos 6x} = \int \sec 6x \, dx$. Using the standard integral formula $\int \sec u \, du = \log |\sec u + 	an u| + c$,we have: $\int \sec 6x \, dx = \frac{1}{6} \log |\sec 6x + 	an 6x| + c$.

$\int \frac{dx}{4\cos^3 2x - 3\cos 2x} = $

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