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

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Question

(B) We have $I = \int x \sin x \sec^3 x \, dx = \int x 	an x \sec^2 x \, dx$. Using integration by parts,let $u = x$ and $dv = 	an x \sec^2 x \, dx$

Vedclass · Accepted Answer

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

Vedclass · Answer

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

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

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