The value of int x sin x sec ^{3} x , dx is | vedclass

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(B) We have $I = \int x \sin x \sec ^{3} x \, dx = \int x 	an x \sec ^{2} x \, dx$.Let $u = x$ and $dv = 	an x \sec ^{2} x \, dx$.Then $du = dx$ and

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$\frac{1}{2}[x \sec ^{2} x-	an x]+c$

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(B) We have $I = \int x \sin x \sec ^{3} x \, dx = \int x 	an x \sec ^{2} x \, dx$.Let $u = x$ and $dv = 	an x \sec ^{2} x \, dx$.Then $du = dx$ and $v = \int 	an x \sec ^{2} x \, dx = \frac{	an ^{2} x}{2}$.Using integration by parts,$\int u \, dv = uv - \int v \, du$:$I = x \cdot \frac{	an ^{2} x}{2} - \int \frac{	an ^{2} x}{2} \, dx$$I = \frac{x 	an ^{2} x}{2} - \frac{1}{2} \int (\sec ^{2} x - 1) \, dx$$I = \frac{x(\sec ^{2} x - 1)}{2} - \frac{1}{2} (	an x - x) + c$$I = \frac{x \sec ^{2} x}{2} - \frac{x}{2} - \frac{	an x}{2} + \frac{x}{2} + c$$I = \frac{1}{2} [x \sec ^{2} x - 	an x] + c$.

The value of $\int x \sin x \sec ^{3} x \, dx$ is

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