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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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} = $

Similar Questions

$\int \frac{d x}{x^{2}+2 x+2}$ equals

$\int \sqrt{1 + \sin \left( \frac{x}{4} \right)} \, dx$

Find the integral of the function $\frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha}$.

$\int {\left( {\cos \frac{x}{2} - \sin \frac{x}{2}} \right)^2} dx = $

If $\frac{d}{d x}(f(x))=4 x^3-\frac{3}{x^4}$ and $f(2)=0$,then $f(x)=$ . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module