int frac{sec^2 x}{1 + tan x} , dx = | vedclass

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

Question

(C) Let $I = \int \frac{\sec^2 x}{1 + 	an x} \, dx$. Substitute $t = 1 + 	an x$. Then,the derivative is $dt = \sec^2 x \, dx$. Substituting these in

Vedclass · Accepted Answer

$\log (1 + 	an x) + c$

Vedclass · Answer

(C) Let $I = \int \frac{\sec^2 x}{1 + 	an x} \, dx$. Substitute $t = 1 + 	an x$. Then,the derivative is $dt = \sec^2 x \, dx$. Substituting these into the integral,we get: $I = \int \frac{1}{t} \, dt$. Integrating with respect to $t$,we obtain: $I = \log |t| + c$. Replacing $t$ with $1 + 	an x$,we get: $I = \log |1 + 	an x| + c$.

$\int \frac{\sec^2 x}{1 + \tan x} \, dx = $

Similar Questions

Integrate the function $\frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}}$.

$\int \frac{1+x \cos x}{x\left[1-x^2\left(e^{\sin x}\right)^2\right]} d x=$

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

$\int \frac{1}{\sqrt{x}} \tan^4(\sqrt{x}) \sec^2(\sqrt{x}) \, dx = $

$\int \frac{dx}{x \log x \log(\log x)} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module