Integral of this form How can this integral be solved as it is of this form:
$$\int \frac{-2x}{\sqrt{1-x^2}} dx$$
Having the derivative of the function inside the square root underneath.
 A: set $$t=1-x^2$$ and we have $$dt=-2xdx$$ and our integral will be $$\int \frac{1}{\sqrt{t}}dt$$
A: 
$$\int  \frac { -2x }{ \sqrt { 1-x^{ 2 } }  } dx=\int  \frac { d\left( 1-x^{ 2 } \right)  }{ \sqrt { 1-x^{ 2 } }  } =2\sqrt { 1-x^{ 2 } } +C$$

A: You already got an answer, but a more general result is this (as far as I know by Charles Hermite):
If $P(x)$ is a polynomial of degree $\geq 1$, then there will exist a polynomial $Q(x)$ of degree $n-1$ and a constant $K$ for which:
$$\int \frac{P(x)}{\sqrt{ax^2+bx+c}} dx = Q(x)\sqrt{ax^2+bx+c} + \int \frac K {\sqrt{ax^2+bx+c}}dx$$
A: $$\int \frac{-2x}{\sqrt{1-x^2}} dx$$
$$x=\sin \theta => dx= \cos \theta\,d\theta$$
$$\int \frac{-2\sin \theta}{\sqrt{1-\sin^2\theta}} \cos \theta d\theta$$
$$\int {-2\sin \theta}{}  d\theta=2\cos\theta \, +C$$
$$\arcsin x=\arcsin\sin\theta=\theta$$
$$2\cos(\arcsin x) \, +C=2\sqrt{1-x^2}+C$$


$$\cos^2 θ + \sin^2 θ = 1\,\,,\, θ = \arcsin(x) $$
$$(\sin(\arcsin x))^2+(\cos(\arcsin x))^2=1$$ $$x^2+(\cos(\arcsin
 x))^2=1$$ $$(\cos(\arcsin x))^2=1-x^2$$ $$\cos(\arcsin
 x)=\sqrt{1-x^2}$$

