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Question
Evaluate the following: `int x.sin^-1 x.dx`
Solution
Let I = `int x.sin^-1 x.dx`
= `int (sin^-1x).xdx`
= `(sin^-1x) int x.dx - int[{d/dx(sin^-1x) int x.dx}].dx`
= `(sin^-1x) (x^2/2) - int (1/sqrt(1 - x^2))(x^2/2).dx`
= `(x^2 sin^-1x)/(2) + (1)/(2) int (-x^2)/sqrt(1 - x^2).dx`
= `(x^2 sin^-1x)/(2) + (1)/(2) int ((1 - x^2) - 1)/sqrt(1 - x^2).dx`
= `(x^2 sin^-1x)/(2) + (1)/(2) int [sqrt(1 - x^2) - (1)/sqrt(1 - x^2)].dx`
= `(x^2 sin^-1x)/(2) + (1)/(2) int sqrt(1 - x^2).dx - (1)/(2) int (1)/sqrt(1 - x^2).dx`
= `(x^2 sin^-1x)/(2) + (1)/(2) [x/2 sqrt(1 - x^2) + 1/2 sin^-1x] - 1/2 sin^-1 x + c`
= `(x^2 sin^-1x)/(2) + (1)/(4) xsqrt(1 - x^2) - (1)/(4) sin^-1 x + c`.
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