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Find ∫sin-1x(1-x2)3/2dx. - Mathematics

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Question

Find `int (sin^-1x)/(1 - x^2)^(3//2) dx`.

Sum
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Solution

Let I = `int (sin^-1x)/(1 - x^2)^(3//2) dx`

Consider t = sin–1 x

`dt/dx = 1/sqrt(1 - x^2)`

∴ I = `int (t.dt)/((1 - x^2))`

= `int (t.dt)/((1 - sin^2t))`

= `int (t.dt)/(cos^2t)`

= `int t . sec^2 t  dt`

On integrating by parts

= `t int sec^2t.dt - int {(d(t))/dt int sec^2 t}dt`

= `t tan t - int 1.tan t  dt`

= t tan t – log sec t + C

= sin–1x tan [sin–1x] – log sec [sin–1x] + C

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2022-2023 (March) Outside Delhi Set 2

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