int frac{sec x , dx}{sqrt{cos 2x}} = | vedclass

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

Question

(A) We have the integral $I = \int \frac{\sec x \, dx}{\sqrt{\cos 2x}}$.Using the identity $\cos 2x = \cos^2 x - \sin^2 x$,we get:$I = \int \frac{\sec

Vedclass · Accepted Answer

$\sin^{-1}(	an x)$

Vedclass · Answer

(A) We have the integral $I = \int \frac{\sec x \, dx}{\sqrt{\cos 2x}}$.Using the identity $\cos 2x = \cos^2 x - \sin^2 x$,we get:$I = \int \frac{\sec x \, dx}{\sqrt{\cos^2 x - \sin^2 x}}$.Multiply the numerator and denominator by $\sec x$:$I = \int \frac{\sec^2 x \, dx}{\sqrt{\frac{\cos^2 x - \sin^2 x}{\cos^2 x}}} = \int \frac{\sec^2 x \, dx}{\sqrt{1 - 	an^2 x}}$.Let $t = 	an x$,then $dt = \sec^2 x \, dx$.Substituting these into the integral:$I = \int \frac{dt}{\sqrt{1 - t^2}} = \sin^{-1}(t) + C = \sin^{-1}(	an x) + C$.

$\int \frac{\sec x \, dx}{\sqrt{\cos 2x}} = $

Similar Questions

$\int \frac{\sin ^6 x}{\cos ^8 x} d x=$

$\int \frac{d x}{\sin \left(x-\frac{\pi}{3}\right) \cos x} = $

$\int \frac{d x}{(x+2) \sqrt{x+1}} = $

$\int \cos ^{3} x \cdot e^{\log (\sin x)} d x$ is equal to

$\int {{e^{3\log x}}{{({x^4} + 1)}^{ - 1}}\,dx} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module