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

Question
Applications of integrals
asked 2020-12-07
Evaluate the integral.
\(\displaystyle\int\frac{{{\sin}^{{-{{1}}}}{x}}}{{\sqrt{{{1}-{x}^{{2}}}}}}\)

Answers (1)

2020-12-08
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.
0

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