int {{{tan }^{ - 1}}sqrt {frac{{1 - cos 2x}}{ | vedclass

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

Question

(C) We know that $1 - \cos 2x = 2\sin^2 x$ and $1 + \cos 2x = 2\cos^2 x$. Substituting these into the integral: $\int {{	an ^{ - 1}}\sqrt {\frac{{2\s

Vedclass · Accepted Answer

$\frac{{{x^2}}}{2} + c$

Vedclass · Answer

(C) We know that $1 - \cos 2x = 2\sin^2 x$ and $1 + \cos 2x = 2\cos^2 x$. Substituting these into the integral: $\int {{	an ^{ - 1}}\sqrt {\frac{{2\sin^2 x}}{{2\cos^2 x}}} } \,dx = \int {{	an ^{ - 1}}\sqrt {	an^2 x} } \,dx$ $= \int {{	an ^{ - 1}}(	an x)\,dx} = \int x \,dx$ $= \frac{{{x^2}}}{2} + c$.

$\int {{{\tan }^{ - 1}}\sqrt {\frac{{1 - \cos 2x}}{{1 + \cos 2x}}} } \;dx = $

Similar Questions

Find the following integral: $\int(4 e^{3 x}+1) d x$

Find the following integral: $\int(ax^{2} + bx + c) dx$

Find the following integral: $\int \frac{dx}{x^{2}-16}$

$\int \frac{1}{7-6 x-x^2} d x=$

$\int \frac{1}{9 \cos ^2 x-24 \sin x \cos x+16 \sin ^2 x} d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module