int sec x tan^3 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 $I = \int \sec x 	an^3 x \, dx$. Rewrite $	an^3 x$ as $	an^2 x \cdot 	an x$: $I = \int \sec x (\sec^2 x - 1)

Vedclass · Accepted Answer

$\frac{1}{3} \sec^3 x - \sec x + c$

Vedclass · Answer

(A) We need to evaluate the integral $I = \int \sec x 	an^3 x \, dx$. Rewrite $	an^3 x$ as $	an^2 x \cdot 	an x$: $I = \int \sec x (\sec^2 x - 1) 	an x \, dx$ $I = \int \sec^2 x (\sec x 	an x) \, dx - \int \sec x 	an x \, dx$ Let $t = \sec x$,then $dt = \sec x 	an x \, dx$. Substituting these into the integral: $I = \int t^2 \, dt - \int 1 \, dt$ $I = \frac{t^3}{3} - t + c$ Substituting back $t = \sec x$: $I = \frac{1}{3} \sec^3 x - \sec x + c$.

$\int \sec x \tan^3 x \, dx = $

Similar Questions

The integral $\int \frac{\left(1-\frac{1}{\sqrt{3}}\right)(\cos x-\sin x)}{\left(1+\frac{2}{\sqrt{3}} \sin 2 x\right)} d x$ is equal to

$\int \frac{(1-4 \sin^2 x) \cos x}{\cos (3x+2)} dx =$

$\int {\frac{{{x^2} - 1}}{{{x^4} + {x^2} + 1}}} \,dx$ is equal to

If $\int \frac{dx}{\sin^3 x + \cos^3 x} = A \log \left|\frac{\sqrt{2}+t}{\sqrt{2}-t}\right| + B \tan^{-1}(t) + c$,then $\left(\frac{B}{A}, t\right) =$

Evaluate the integral: $\int \sqrt{4x(4x + 4)} \, dx$

Vedclass Test Series

Exam Paper Generator

Online Exam Module