Mathematics

Mathematics

Problem

∫sin⁡−1x(1−x2)3/2 dx=\int \frac{\sin^{-1}x}{(1-x^2)^{3/2}}\,dx = (a) x1−x2sin⁡−1x+12log⁡(1−x2)+c\frac{x}{\sqrt{1-x^2}}\sin^{-1}x + \frac{1}{2}\log(1-x^2) + c (b) x1−x2sin⁡−1x−12log⁡(1−x2)+c\frac{x}{\sqrt{1-x^2}}\sin^{-1}x - \frac{1}{2}\log(1-x^2) + c (c) 11−x2sin⁡−1x−12log⁡(1−x2)+c\frac{1}{\sqrt{1-x^2}}\sin^{-1}x - \frac{1}{2}\log(1-x^2) + c (d) 11−x2sin⁡−1x+12log⁡(1−x2)+c\frac{1}{\sqrt{1-x^2}}\sin^{-1}x + \frac{1}{2}\log(1-x^2) + c (a) x1−x2sin⁡−1x+12log⁡(1−x2)+c\frac{x}{\sqrt{1-x^2}}\sin^{-1}x + \frac{1}{2}\log(1-x^2) + c (b) x1−x2sin⁡−1x−12log⁡(1−x2)+c\frac{x}{\sqrt{1-x^2}}\sin^{-1}x - \frac{1}{2}\log(1-x^2) + c (c) 11−x2sin⁡−1x−12log⁡(1−x2)+c\frac{1}{\sqrt{1-x^2}}\sin^{-1}x - \frac{1}{2}\log(1-x^2) + c (d) 11−x2sin⁡−1x+12log⁡(1−x2)+c\frac{1}{\sqrt{1-x^2}}\sin^{-1}x + \frac{1}{2}\log(1-x^2) + c
Intuitive explanation

This integral pairs an inverse trig function with a denominator that's a derivative signature—the key is integration by parts. Notice that ddx(sin⁡−1x)=11−x2\frac{d}{dx}(\sin^{-1}x) = \frac{1}{\sqrt{1-x^2}} and (1−x2)−3/2(1-x^2)^{-3/2} is related to the derivative of x1−x2\frac{x}{\sqrt{1-x^2}}. Choose u=sin⁡−1xu = \sin^{-1}x and dv=(1−x2)−3/2dxdv = (1-x^2)^{-3/2}dx, then carefully compute dudu and vv to reduce the problem to a simpler integral you can handle.

Intuitive explanation

This integral pairs an inverse trig function with a denominator that's a derivative signature—the key is integration by parts. Notice that ddx(sin⁡−1x)=11−x2\frac{d}{dx}(\sin^{-1}x) = \frac{1}{\sqrt{1-x^2}} and (1−x2)−3/2(1-x^2)^{-3/2} is related to the derivative of x1−x2\frac{x}{\sqrt{1-x^2}}. Choose u=sin⁡−1xu = \sin^{-1}x and dv=(1−x2)−3/2dxdv = (1-x^2)^{-3/2}dx, then carefully compute dudu and vv to reduce the problem to a simpler integral you can handle.

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