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(B) Let $I = \int 	an^4 x \, dx$. We can write $	an^4 x$ as $	an^2 x \cdot 	an^2 x$. Using the identity $	an^2 x = \sec^2 x - 1$,we get: $I = \in

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $I = \int 	an^4 x \, dx$. We can write $	an^4 x$ as $	an^2 x \cdot 	an^2 x$. Using the identity $	an^2 x = \sec^2 x - 1$,we get: $I = \int 	an^2 x (\sec^2 x - 1) \, dx$ $I = \int 	an^2 x \sec^2 x \, dx - \int 	an^2 x \, dx$ $I = \int 	an^2 x \sec^2 x \, dx - \int (\sec^2 x - 1) \, dx$ For the first integral,let $u = 	an x$,then $du = \sec^2 x \, dx$. $I = \int u^2 \, du - (	an x - x) + c$ $I = \frac{u^3}{3} - 	an x + x + c$ Substituting $u = 	an x$ back,we get: $I = \frac{1}{3} 	an^3 x - 	an x + x + c$.

$\int \tan^4 x \, dx = $

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