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Evaluate the following : ∫x.sin-1x.dx

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Question

Evaluate the following: `int x.sin^-1 x.dx`

Sum
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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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Chapter 3: Indefinite Integration - Exercise 3.3 [Page 137]

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