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(D) To evaluate the integral $\int x \sec^2 x \, dx$,we use the method of integration by parts: $\int u \, dv = uv - \int v \, du$.Let $u = x$ and $dv

Vedclass · Accepted Answer

$x 	an x + \log \cos x + c$

Vedclass · Answer

(D) To evaluate the integral $\int x \sec^2 x \, dx$,we use the method of integration by parts: $\int u \, dv = uv - \int v \, du$.Let $u = x$ and $dv = \sec^2 x \, dx$.Then $du = dx$ and $v = 	an x$.Applying the formula:$\int x \sec^2 x \, dx = x 	an x - \int 	an x \, dx$.Since $\int 	an x \, dx = \log |\sec x| = -\log |\cos x|$,we have:$\int x \sec^2 x \, dx = x 	an x - (-\log |\cos x|) + c = x 	an x + \log |\cos x| + c$.

$\int x \sec^2 x \, dx = $

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