int sec x tan^3 x , dx = | vedclass

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Question

(A) We need to evaluate the integral $I = \int \sec x 	an^3 x \, dx$. Rewrite $	an^3 x$ as $	an^2 x \cdot 	an x$: $I = \int \sec x (\sec^2 x - 1)

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

Vedclass · Answer

(A) We need to evaluate the integral $I = \int \sec x 	an^3 x \, dx$. Rewrite $	an^3 x$ as $	an^2 x \cdot 	an x$: $I = \int \sec x (\sec^2 x - 1) 	an x \, dx$ $I = \int \sec^2 x (\sec x 	an x) \, dx - \int \sec x 	an x \, dx$ Let $t = \sec x$,then $dt = \sec x 	an x \, dx$. Substituting these into the integral: $I = \int t^2 \, dt - \int 1 \, dt$ $I = \frac{t^3}{3} - t + c$ Substituting back $t = \sec x$: $I = \frac{1}{3} \sec^3 x - \sec x + c$.

$\int \sec x \tan^3 x \, dx = $

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