int tan ^8 x cdot sec ^4 x , dx = . . . . . | vedclass

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let $I = \int 	an ^8 x \cdot \sec ^4 x \, dx$. We can write $\sec ^4 x$ 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 back $u = 	an x$: $I = \frac{	an ^9 x}{9} + \frac{	an ^{11} x}{11} + C$. Comparing this with the given options,the correct option is $D$.

$\int \tan ^8 x \cdot \sec ^4 x \, dx = $ . . . . . . $+ C$.

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