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

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

Question

(D) Let $I = \int {\frac{{1 + {{	an }^2}x}}{{1 - {{	an }^2}x}}dx} $.Using the identity $1 + 	an^2 x = \sec^2 x$,we get $I = \int {\frac{{{{\sec }^2

Vedclass · Accepted Answer

$\frac{1}{2}\log \left( {\frac{{1 + 	an x}}{{1 - 	an x}}} ight) + c$

Vedclass · Answer

(D) Let $I = \int {\frac{{1 + {{	an }^2}x}}{{1 - {{	an }^2}x}}dx} $.Using the identity $1 + 	an^2 x = \sec^2 x$,we get $I = \int {\frac{{{{\sec }^2}x}}{{1 - {{	an }^2}x}}dx} $.Substitute $	an x = t$,then $\sec^2 x \, dx = dt$.The integral becomes $I = \int {\frac{{dt}}{{1 - {t^2}}}} $.Using the standard formula $\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \log \left| \frac{a+x}{a-x} ight| + c$,we have:$I = \frac{1}{2(1)} \log \left| \frac{1+t}{1-t} ight| + c$.Substituting $t = 	an x$ back,we get $I = \frac{1}{2} \log \left| \frac{1 + 	an x}{1 - 	an x} ight| + c$.

$\int {\frac{{1 + {{\tan }^2}x}}{{1 - {{\tan }^2}x}}\,dx} $ is equal to

Similar Questions

The integral of $\sqrt{1 + 2 \cot x (\cot x + \csc x)}$ with respect to $x$ is:

If $x > 0$ and $x \neq (2n+1) \frac{\pi}{2}$,then $\int \left(x \sqrt{x} - e^{\log(\sec x \tan x)} + \frac{3x^2 - 2x + 1}{x^2}\right) dx =$

Find the integral of the function $\cos^{4} 2x$.

$\int \frac{1+x+\sqrt{x+x^2}}{\sqrt{x}+\sqrt{1+x}} d x$ is equal to

$\int \frac{dx}{\cos x - \sin x}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module