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(B) To evaluate the integral $\int 	an^4 x \, dx$,we use the identity $	an^2 x = \sec^2 x - 1$. $\int 	an^4 x \, dx = \int 	an^2 x (	an^2 x) \, d

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) To evaluate the integral $\int 	an^4 x \, dx$,we use the identity $	an^2 x = \sec^2 x - 1$. $\int 	an^4 x \, dx = \int 	an^2 x (	an^2 x) \, dx$ $= \int 	an^2 x (\sec^2 x - 1) \, dx$ $= \int 	an^2 x \sec^2 x \, dx - \int 	an^2 x \, dx$ $= \int 	an^2 x \sec^2 x \, dx - \int (\sec^2 x - 1) \, dx$ For the first part,let $u = 	an x$,then $du = \sec^2 x \, dx$. $\int u^2 \, du = \frac{u^3}{3} = \frac{	an^3 x}{3}$. For the second part,$\int (\sec^2 x - 1) \, dx = 	an x - x$. Combining these,we get: $\frac{1}{3} 	an^3 x - (	an x - x) + c = \frac{1}{3} 	an^3 x - 	an x + x + c$.

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

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