Evaluate the integral: int tan ^8 x sec ^4 x | vedclass

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

Question

(C) Let $I = \int 	an ^8 x \sec ^4 x \, dx$. We can rewrite $\sec ^4 x$ as $\sec ^2 x \cdot \sec ^2 x$. So,$I = \int 	an ^8 x \cdot \sec ^2 x \cdot

Vedclass · Accepted Answer

$\frac{	an ^{11} x}{11} + \frac{	an ^9 x}{9} + c$

Vedclass · Answer

(C) Let $I = \int 	an ^8 x \sec ^4 x \, dx$. We can rewrite $\sec ^4 x$ as $\sec ^2 x \cdot \sec ^2 x$. So,$I = \int 	an ^8 x \cdot \sec ^2 x \cdot \sec ^2 x \, dx$. Since $\sec ^2 x = 1 + 	an ^2 x$,we have: $I = \int 	an ^8 x (1 + 	an ^2 x) \sec ^2 x \, dx$. Let $u = 	an x$,then $du = \sec ^2 x \, dx$. Substituting these into the integral: $I = \int u^8 (1 + u^2) \, du = \int (u^8 + u^{10}) \, du$. Integrating with respect to $u$: $I = \frac{u^9}{9} + \frac{u^{11}}{11} + c$. Substituting $u = 	an x$ back: $I = \frac{	an ^9 x}{9} + \frac{	an ^{11} x}{11} + c$.

Evaluate the integral: $\int \tan ^8 x \sec ^4 x \, dx$.

Similar Questions

The integral $\int \frac{\sin ^2 x \cos ^2 x \,dx}{(\sin ^5 x+\cos ^3 x \sin ^2 x+\sin ^3 x \cos ^2 x+\cos ^5 x)^2}$ is equal to (where $C$ is a constant of integration).

$\int \frac{e^x(x + 1)}{\cos^2(x e^x)} dx = $

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

$\int \frac{d x}{x \ln (x) \ln ^2(x) \ln ^3(x) \ldots \ln ^m(x)}=\frac{(\ln (x))^K}{K}+C$
$\Rightarrow 2 K=$

Integrate the function $\frac{x^{3} \sin \left(\tan ^{-1} x^{4}\right)}{1+x^{8}}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module