# Evaluate the integrals. \int \frac{\sin^{-1}x}{\sqrt{1-x^{2}}}dx

Question
Applications of integrals
Evaluate the integrals.
$$\displaystyle\int{\frac{{{{\sin}^{{-{1}}}{x}}}}{{\sqrt{{{1}-{x}^{{{2}}}}}}}}{\left.{d}{x}\right.}$$

2021-02-13
Step 1
Consider the provided integral,
$$\displaystyle\int{\frac{{{{\sin}^{{-{1}}}{x}}}}{{\sqrt{{{1}-{x}^{{{2}}}}}}}}{\left.{d}{x}\right.}$$
Evaluate the integrals.
Apply the Substitution method,
$$\displaystyle{u}={{\sin}^{{-{1}}}{x}}\Rightarrow{d}{u}={\frac{{{1}}}{{\sqrt{{{1}-{x}^{{{2}}}}}}}}{\left.{d}{x}\right.}$$
Step 2
Therefore,
$$\displaystyle\int{\frac{{{{\sin}^{{-{1}}}{x}}}}{{\sqrt{{{1}-{x}^{{{2}}}}}}}}{\left.{d}{x}\right.}=\int{u}{d}{u}$$
$$\displaystyle={\frac{{{u}^{{{1}+{1}}}}}{{{1}+{1}}}}+{C}$$
$$\displaystyle={\frac{{{u}^{{{2}}}}}{{{2}}}}+{C}$$
Substitute back $$\displaystyle{u}={{\sin}^{{-{1}}}{x}}$$.
$$\displaystyle\int{\frac{{{{\sin}^{{-{1}}}{x}}}}{{\sqrt{{{1}-{x}^{{{2}}}}}}}}{\left.{d}{x}\right.}={\frac{{{\left({{\sin}^{{-{1}}}{x}}\right)}^{{{2}}}}}{{{2}}}}+{C}$$
$$\displaystyle={\frac{{{1}}}{{{2}}}}{\left({{\sin}^{{-{1}}}{x}}\right)}^{{{2}}}+{C}$$
Hence.

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