int frac{sin (tan ^{-1} x)}{1+x^2} d x= . . | vedclass

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(A) To evaluate the integral $I = \int \frac{\sin (	an ^{-1} x)}{1+x^2} d x$,we use the method of substitution.Let $u = 	an ^{-1} x$.Then,differenti

Vedclass · Accepted Answer

$-\cos (	an ^{-1} x)$

Vedclass · Answer

(A) To evaluate the integral $I = \int \frac{\sin (	an ^{-1} x)}{1+x^2} d x$,we use the method of substitution.Let $u = 	an ^{-1} x$.Then,differentiating both sides with respect to $x$,we get $du = \frac{1}{1+x^2} dx$.Substituting these into the integral,we have $I = \int \sin(u) du$.The integral of $\sin(u)$ is $-\cos(u) + C$.Substituting back $u = 	an ^{-1} x$,we get $I = -\cos (	an ^{-1} x) + C$.

$\int \frac{\sin (\tan ^{-1} x)}{1+x^2} d x=$ . . . . . . $+C$.

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