Integrate the following function with respect | vedclass

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Question

(A) Let $I = \int \frac{\sin(	an^{-1} x)}{1+x^2} dx$. We know that the derivative of $	an^{-1} x$ is $\frac{1}{1+x^2}$. Let $t = 	an^{-1} x$. Then,

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

Vedclass · Answer

(A) Let $I = \int \frac{\sin(	an^{-1} x)}{1+x^2} dx$. We know that the derivative of $	an^{-1} x$ is $\frac{1}{1+x^2}$. Let $t = 	an^{-1} x$. Then,$dt = \frac{1}{1+x^2} dx$. Substituting these into the integral,we get: $I = \int \sin(t) dt$. Integrating $\sin(t)$ with respect to $t$,we get: $I = -\cos(t) + C$. Substituting back $t = 	an^{-1} x$,we get: $I = -\cos(	an^{-1} x) + C$.

Integrate the following function with respect to $x$: $\frac{\sin(\tan^{-1} x)}{1+x^2}$

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