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