int x(tan^2 x) dx = | vedclass

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

Question

(A) We need to evaluate the integral $I = \int x 	an^2 x dx$.Using the identity $	an^2 x = \sec^2 x - 1$,we get:$I = \int x(\sec^2 x - 1) dx = \int

Vedclass · Accepted Answer

$x 	an x - \log_e(\sec x) - \frac{x^2}{2} + C$

Vedclass · Answer

(A) We need to evaluate the integral $I = \int x 	an^2 x dx$.Using the identity $	an^2 x = \sec^2 x - 1$,we get:$I = \int x(\sec^2 x - 1) dx = \int x \sec^2 x dx - \int x dx$.Applying integration by parts to $\int x \sec^2 x dx$ (taking $u = x$ and $dv = \sec^2 x dx$):$I = [x 	an x - \int 1 \cdot 	an x dx] - \frac{x^2}{2} + C$.Since $\int 	an x dx = \log_e(\sec x)$:$I = x 	an x - \log_e(\sec x) - \frac{x^2}{2} + C$.Thus,the correct option is $A$.

$\int x(\tan^2 x) dx =$

Similar Questions

Evaluate the integral: $\int \left(\frac{8^{1+x}+4^{1+x}}{2^{2x}}\right) dx$

$\int \frac{\tan x}{\sec x + \tan x} \, dx = $

$\int(\sqrt{1-\sin x}+\sqrt{1+\sin x}) dx =f(x)+c$,where $c$ is the constant of integration. If $\frac{5 \pi}{2}$

$\int \frac{d x}{\sin ^{2} x \cos ^{2} x}$ equals

If $f(x) = \frac{x}{x+1}, x \neq -1$ and $(fof)(x) = F(x)$,then $\int F(x) \, dx$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module