int {xsin x {{sec }^3} x,dx} is equal to | vedclass

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(B) We need to evaluate the integral $I = \int x \sin x \sec^3 x \, dx$. First,simplify the integrand: $I = \int x 	an x \sec^2 x \, dx$. Using integ

Vedclass · Accepted Answer

$\frac{1}{2}\left( {x\ {{\sec }^2}\ x - 	an x} ight) + C$

Vedclass · Answer

(B) We need to evaluate the integral $I = \int x \sin x \sec^3 x \, dx$. First,simplify the integrand: $I = \int x 	an x \sec^2 x \, dx$. 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 $\int u \, dv = uv - \int v \, du$: $I = x \cdot \frac{	an^2 x}{2} - \int \frac{	an^2 x}{2} \, dx$. $I = \frac{x 	an^2 x}{2} - \frac{1}{2} \int (\sec^2 x - 1) \, dx$. $I = \frac{x 	an^2 x}{2} - \frac{1}{2} (	an x - x) + C$. $I = \frac{x 	an^2 x}{2} - \frac{1}{2} 	an x + \frac{x}{2} + C$. Since $	an^2 x = \sec^2 x - 1$: $I = \frac{x(\sec^2 x - 1)}{2} - \frac{1}{2} 	an x + \frac{x}{2} + C$. $I = \frac{x \sec^2 x}{2} - \frac{x}{2} - \frac{1}{2} 	an x + \frac{x}{2} + C$. $I = \frac{1}{2} (x \sec^2 x - 	an x) + C$.

$\int {x\sin x\ {{\sec }^3}\ x\,dx} $ is equal to

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