int frac{dx}{cos 2x + sin^2 x} = | vedclass

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

Question

(B) We are given the integral $I = \int \frac{dx}{\cos 2x + \sin^2 x}$.Using the trigonometric identity $\cos 2x = 1 - 2\sin^2 x$,we substitute this i

Vedclass · Accepted Answer

$	an x + c$

Vedclass · Answer

(B) We are given the integral $I = \int \frac{dx}{\cos 2x + \sin^2 x}$.Using the trigonometric identity $\cos 2x = 1 - 2\sin^2 x$,we substitute this into the denominator:$I = \int \frac{dx}{1 - 2\sin^2 x + \sin^2 x}$$I = \int \frac{dx}{1 - \sin^2 x}$Using the identity $1 - \sin^2 x = \cos^2 x$,we get:$I = \int \frac{dx}{\cos^2 x}$$I = \int \sec^2 x \, dx$The integral of $\sec^2 x$ is $	an x + c$.Therefore,$I = 	an x + c$.

$\int \frac{dx}{\cos 2x + \sin^2 x} = $

Similar Questions

If $f^{\prime}(x)=k(\cos x+\sin x)$ and $f(0)=9, f\left(\frac{\pi}{2}\right)=15$,then $f(x)=$

$ \int \frac{1}{1+e^{x}} d x $ is equal to

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

If $\int (\sin 2x - \cos 2x) \,dx = \frac{1}{\sqrt{2}} \sin(2x - a) + b$,then

Find: $\int \sin 2x \cos 3x \, dx$

Vedclass Test Series

Exam Paper Generator

Online Exam Module