int frac{sin ^{-1} x}{sqrt{1-x^2}} d x is equ | vedclass

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(B) To evaluate the integral $I = \int \frac{\sin ^{-1} x}{\sqrt{1-x^2}} d x$, we use the method of substitution.Let $t = \sin ^{-1} x$.Then, differen

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$\frac{1}{2}\left(\sin ^{-1} x\right)^2+c$

Vedclass · Answer

(B) To evaluate the integral $I = \int \frac{\sin ^{-1} x}{\sqrt{1-x^2}} d x$, we use the method of substitution.Let $t = \sin ^{-1} x$.Then, differentiating both sides with respect to $x$, we get $\frac{dt}{dx} = \frac{1}{\sqrt{1-x^2}}$, which implies $dt = \frac{1}{\sqrt{1-x^2}} dx$.Substituting these into the integral, we get $I = \int t dt$.Integrating with respect to $t$, we obtain $I = \frac{1}{2} t^2 + c$.Finally, substituting $t = \sin ^{-1} x$ back into the expression, we get $I = \frac{1}{2} (\sin ^{-1} x)^2 + c$.

$\int \frac{\sin ^{-1} x}{\sqrt{1-x^2}} d x$ is equal to, where $c$ is an arbitrary constant.

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