Evaluate
$$
\int\frac{\mathrm{d}x}{(1+x^2)\sqrt{1-x^2}}.
$$
Set $x=\tan u\implies\mathrm{d}x=\sec^2u\,\mathrm{d}u$.
$$
I=\int\frac{\mathrm{d}x}{(1+x^2)\sqrt{1-x^2}}=\int\frac{\sec^2u\,\mathrm{d}u}{\sec^2u\,\dfrac{\sqrt{1-2\sin^2u}}{\cos u}}=\int\frac{\cos u\,\mathrm{d}u}{\sqrt{1-2\sin^2u}}.
$$
Set $t=\sqrt2\sin u\implies\mathrm{d}t=\sqrt2\cos u\,\mathrm{d}u$
$$
I=\frac1{\sqrt2}\int\frac{\mathrm{d}t}{\sqrt{1-t^2}}=\frac1{\sqrt2}\sin^{-1}t+c=\frac1{\sqrt2}\sin^{-1}\left(\sqrt{2}\sin u\right)+c=\frac1{\sqrt2}\sin^{-1}\left(\frac{\sqrt2x}{\sqrt{1+x^2}}\right)+c.
$$
But my reference gives the solution $\dfrac{-1}{\sqrt{2}}\sin^{-1}\sqrt{\dfrac{1-x^2}{1+x^2}}+c$, where am I going wrong with my attempt ?
Best Answer
Set $x=\sin y,-\dfrac\pi2<x<\dfrac\pi2,\tan y=\dfrac x{\sqrt{1-x^2}}$
$$\int\dfrac{dx}{(1+x^2)\sqrt{1-x^2}}=\int\dfrac{dy}{1+\sin^2y}=\dfrac12\int\dfrac{\sec^2y\ dy}{1/2+\tan^2y}=\dfrac1{\sqrt2}\arctan(\sqrt2\tan y)$$
$$=\dfrac1{\sqrt2}\arctan\dfrac{\sqrt2x}{\sqrt{1-x^2}}$$
If $\arctan\dfrac{\sqrt2x}{\sqrt{1-x^2}}=u,-\dfrac\pi2<u<\dfrac\pi2$
$\tan u=\dfrac{\sqrt2x}{\sqrt{1-x^2}}$
$\sec u=+\sqrt{1+\dfrac{2x^2}{1-x^2}}=\sqrt{\dfrac{1+x^2}{1-x^2}}$
$\sin u=\dfrac{\tan u}{\sec u}=?$
So, your solution is correct