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What is \int\sin^{-1}(x)dx value?

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To find- $\int \sin^{-1}(x) \, dx$

Solution-  Let $u = \sin^{-1}(x)$, so that $du = \frac{1}{\sqrt{1 - x^2}} \, dx$, and $dv = dx$, so that $v = x$. 
Using the integration by parts formula $\int u \, dv = uv - \int v \, du$, we get:
$\int \sin^{-1}(x) \, dx = x \sin^{-1}(x) - \int \frac{x}{\sqrt{1 - x^2}} \, dx$

Now, solve $\int \frac{x}{\sqrt{1 - x^2}} \, dx$ using the substitution $w = 1 - x^2$, so that $dw = -2x \, dx$ or $x \, dx = -\frac{dw}{2}$. Substituting, we get:
$\int \frac{x}{\sqrt{1 - x^2}} \, dx = -\frac{1}{2} \int w^{-1/2} \, dw = -\sqrt{w} = -\sqrt{1 - x^2}$

Thus, the final answer is:
$\int \sin^{-1}(x) \, dx = x \sin^{-1}(x) + \sqrt{1 - x^2} + C$

Posted by

Saniya Khatri

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