Question:
Integrate the function: \((sin^{-1}x)^2\)
Integrate the function: \((sin^{-1}x)^2\)
Updated On: Oct 4, 2023
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Solution and Explanation
The correct answer is: \(=x(sin^{-1}x)^2+2\sqrt{1-x^2}sin^{-1}x-2x+C\)
Let \(I=∫(Sin^{-1}x)^2.1 dx\)
Taking\((sin^{-1}x)^2\) as first function and 1 as second function and integrating by parts,we
obtain
\(I=(sin^{-1}x)∫1dx-∫[{\frac{d}{dx}(sin^{-1}x)^2.∫1.dx}]dx\)
\(=(sin^{-1}x)^2.x-∫\frac{2sin^{-1}x}{\sqrt{1-x^2}}.x dx\)
\(=x(sin^{-1}x)^2+∫sin^{-1}x.(\frac{-2x}{\sqrt{1-x^2}})dx\)
\(=x(sin^{-1}x)^2+[sin^{-1}x∫\frac{-2x}{\sqrt{1-x^2}} dx-∫{(\frac{d}{dx} sin^{-1}x)∫\frac{-2x}{\sqrt{1-x^2}} dx}]dx\)
\(=x(sin^{-1}x)^2+[sin^{-1}x.2\sqrt{1-x^2}-∫\frac{1}{\sqrt{1-x^2}}.2\sqrt{1-x^2} dx]\)
\(=x(sin^{-1}x)^2+2\sqrt{1-x^2} sin^{-1}x-∫2.dx\)
\(=x(sin^{-1}x)^2+2\sqrt{1-x^2}sin^{-1}x-2x+C\)
Let \(I=∫(Sin^{-1}x)^2.1 dx\)
Taking\((sin^{-1}x)^2\) as first function and 1 as second function and integrating by parts,we
obtain
\(I=(sin^{-1}x)∫1dx-∫[{\frac{d}{dx}(sin^{-1}x)^2.∫1.dx}]dx\)
\(=(sin^{-1}x)^2.x-∫\frac{2sin^{-1}x}{\sqrt{1-x^2}}.x dx\)
\(=x(sin^{-1}x)^2+∫sin^{-1}x.(\frac{-2x}{\sqrt{1-x^2}})dx\)
\(=x(sin^{-1}x)^2+[sin^{-1}x∫\frac{-2x}{\sqrt{1-x^2}} dx-∫{(\frac{d}{dx} sin^{-1}x)∫\frac{-2x}{\sqrt{1-x^2}} dx}]dx\)
\(=x(sin^{-1}x)^2+[sin^{-1}x.2\sqrt{1-x^2}-∫\frac{1}{\sqrt{1-x^2}}.2\sqrt{1-x^2} dx]\)
\(=x(sin^{-1}x)^2+2\sqrt{1-x^2} sin^{-1}x-∫2.dx\)
\(=x(sin^{-1}x)^2+2\sqrt{1-x^2}sin^{-1}x-2x+C\)
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Concepts Used:
Integration by Partial Fractions
The number of formulas used to decompose the given improper rational functions is given below. By using the given expressions, we can quickly write the integrand as a sum of proper rational functions.
For examples,
