int x(tan^2 x) dx = | vedclass

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Question

(A) We need to evaluate the integral $I = \int x 	an^2 x dx$.Using the identity $	an^2 x = \sec^2 x - 1$,we get:$I = \int x(\sec^2 x - 1) dx = \int

Vedclass · Accepted Answer

$x 	an x - \log_e(\sec x) - \frac{x^2}{2} + C$

Vedclass · Answer

(A) We need to evaluate the integral $I = \int x 	an^2 x dx$.Using the identity $	an^2 x = \sec^2 x - 1$,we get:$I = \int x(\sec^2 x - 1) dx = \int x \sec^2 x dx - \int x dx$.Applying integration by parts to $\int x \sec^2 x dx$ (taking $u = x$ and $dv = \sec^2 x dx$):$I = [x 	an x - \int 1 \cdot 	an x dx] - \frac{x^2}{2} + C$.Since $\int 	an x dx = \log_e(\sec x)$:$I = x 	an x - \log_e(\sec x) - \frac{x^2}{2} + C$.Thus,the correct option is $A$.

$\int x(\tan^2 x) dx =$

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