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

If $\frac{d}{d x} f(x)=4 x^3-\frac{3}{x^4}$ such that $f(2)=0$,then $f(x)$ is equal to

If $f\left( \frac{x - 4}{x + 2} \right) = 2x + 1$ for $x \in R \setminus \{ -2 \}$,then $\int f(x) \,dx$ is equal to (where $C$ is a constant of integration)

The value of $\int \frac{2 x^3-1}{x^4+x} \,d x$ is equal to (where $C$ is a constant of integration.)

$\int \frac{x}{x + 2} dx$ is equal to

$\int \frac{x - 1}{(x + 1)^2} \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module