Evaluate the integral. int sin^-1x/(sqrt(1-x^2))

Evaluate the integral.
$\int \frac{{\mathrm{sin}}^{-1}x}{\sqrt{1-{x}^{2}}}$
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Faiza Fuller
Step 1
Consider the provided integral,
$\int \frac{{\mathrm{sin}}^{-1}x}{\sqrt{1-{x}^{2}}}dx$
Evaluate the integrals.
Apply the Substitution method,
$u={\mathrm{sin}}^{-1}x⇒du=\frac{1}{\sqrt{1-{x}^{2}}}dx$
Step 2
Therefore,
$\int \frac{{\mathrm{sin}}^{-1}x}{\sqrt{1-{x}^{2}}}dx=\int udu$
$=\frac{{u}^{1+1}}{1+1}+C$
$=\frac{{u}^{2}}{2}+C$
Substitute back $u={\mathrm{sin}}^{-1}x$.
$\int \frac{{\mathrm{sin}}^{-1}x}{\sqrt{1-{x}^{2}}}dx=\frac{{\left({\mathrm{sin}}^{-1}x\right)}^{2}}{2}+C$
$=\frac{1}{2}{\left({\mathrm{sin}}^{-1}x\right)}^{2}+C$
Hence.