int sec^4 x tan x ; dx = | vedclass

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

Question

(A) To solve the integral $\int \sec^4 x 	an x \; dx$,we can rewrite the integrand as: $\int \sec^3 x (\sec x 	an x) \; dx$ Let $t = \sec x$. Then,t

Vedclass · Accepted Answer

$\frac{1}{4} \sec^4 x + c$

Vedclass · Answer

(A) To solve the integral $\int \sec^4 x 	an x \; dx$,we can rewrite the integrand as: $\int \sec^3 x (\sec x 	an x) \; dx$ Let $t = \sec x$. Then,the derivative is $dt = \sec x 	an x \; dx$. Substituting these into the integral,we get: $\int t^3 \; dt$ Integrating with respect to $t$,we obtain: $\frac{t^4}{4} + c$ Finally,substituting $t = \sec x$ back into the expression,we get: $\frac{1}{4} \sec^4 x + c$

$\int \sec^4 x \tan x \; dx = $

Similar Questions

The value of $I=\int \frac{x^2}{(a+b x)^2} \,d x$ is

If $\int \frac{2 \sin 2x - 3 \cos x}{2 \sin^2 x - 3 \sin x + 4} dx = f(x) + c$ where $c$ is the constant of integration, then $f\left(\frac{\pi}{2}\right) - f(0) =$

$\int \frac{1}{x+x \log x} d x=$ . . . . . . .

$\int \frac{\cos^3(x)}{\sin^2(x)+\sin(x)} \, dx =$

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module