int e^x tan x(1+tan 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 $I = \int e^x (	an x + 	an^2 x) \, dx$. Recall the identity $1 + 	an^2 x = \sec^2 x$,which implies $	an^2 x =

Vedclass · Accepted Answer

$e^x(	an x - 1)$

Vedclass · Answer

(A) We need to evaluate the integral $I = \int e^x (	an x + 	an^2 x) \, dx$. Recall the identity $1 + 	an^2 x = \sec^2 x$,which implies $	an^2 x = \sec^2 x - 1$. Substituting this into the integral,we get: $I = \int e^x (	an x + \sec^2 x - 1) \, dx$. Rearranging the terms: $I = \int e^x ((	an x - 1) + \sec^2 x) \, dx$. Let $f(x) = 	an x - 1$. Then,$f'(x) = \sec^2 x$. Using the standard integration formula $\int e^x (f(x) + f'(x)) \, dx = e^x f(x) + C$,we have: $I = e^x (	an x - 1) + C$. Therefore,the correct option is $A$.

$\int e^x \tan x(1+\tan x) \, dx = $ . . . . . . $+ C$.

Similar Questions

The integral $\int_{1}^{2} e^{x} \cdot x^{x}(1 + \log_{e} x + 1) dx$ is equal to:

$\int \frac{x-3}{(x-1)^3} e^x \, dx =$

If $\int {{e^{\sec x}}\left( {\sec x + \tan x f(x) + (\sec x \tan x + \sec^2 x)} \right)dx = {e^{\sec x}}f(x) + C}$,then a possible choice of $f(x)$ is

The value of $\int e^{x}\left[\frac{1+\sin x}{1+\cos x}\right] d x$ is equal to

$\int\limits_1^2 {{e^{2x}}} \left( {\frac{1}{x} - \frac{1}{{2{x^2}}}} \right)\,dx$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module