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

$\int \frac{e^x \, dx}{\sqrt{1 - e^{2x}}} = $

Evaluate $\int \sqrt{e^{4x} + e^{2x}} \, dx$.

$\int \frac{(3 x-2) \tan \left(\sqrt{9 x^2-12 x+1}\right)}{\sqrt{9 x^2-12 x+1}} d x=$

$\int \frac{10^{\frac{x}{2}}}{\sqrt{10^{-x}-10^x}} dx=$

$\int 3^{3^x} \cdot 3^x \, dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module