int sec^p x tan x , dx = | vedclass

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(B) Let $I = \int \sec^p x 	an x \, dx$. Substitute $t = \sec x$. Then,$dt = \sec x 	an x \, dx$,which implies $	an x \, dx = \frac{dt}{\sec x} = \

Vedclass · Accepted Answer

$\frac{\sec^p x}{p} + c$

Vedclass · Answer

(B) Let $I = \int \sec^p x 	an x \, dx$. Substitute $t = \sec x$. Then,$dt = \sec x 	an x \, dx$,which implies $	an x \, dx = \frac{dt}{\sec x} = \frac{dt}{t}$. Substituting these into the integral: $I = \int t^p \cdot \frac{dt}{t} = \int t^{p-1} \, dt$. Integrating with respect to $t$: $I = \frac{t^{(p-1)+1}}{(p-1)+1} + c = \frac{t^p}{p} + c$. Replacing $t$ with $\sec x$: $I = \frac{\sec^p x}{p} + c$.

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

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