$Integrate $\int x^2(1-x^2)^{-\frac{3}{2}}dx$$

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I am tired of trying to evaluate
$$\int x^2(1-x^2)^{-\frac{3}{2}}dx.$$
I am substituting $x=\sin \theta$, but get stuck in the remaining calculations. A detailed answer would be greatly appreciated.

Calculus Integrals Trigonometric Functions

Karl Sch

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Let $x=\sin \theta$. Then $dx=\cos \theta d\theta $. So
\[\int x^2(1-x^2)^{3/2}dx=\int \sin^2 \theta(1-\sin ^2 \theta)^{-3/2}\cos \theta d \theta\]
\[=\int \sin^2 \theta\cos ^{-3} \theta \cos \theta d \theta=\int \frac{\sin^2 \theta}{\cos ^2 \theta} d\theta=\int \tan^2 \theta d \theta \]
\[=\int \sec ^2 \theta -1 d \theta= \tan \theta -\theta+c\]
\[=\frac{\sin \theta}{\cos \theta} -\theta +c=\frac{x}{\sqrt{1-x^2}}-\arcsin x+c.\]

Nirenberg

$Integrate $\int x^2(1-x^2)^{-\frac{3}{2}}dx$$

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