Integral of this form

Question

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.

Answer 1

set $$t=1-x^2$$ and we have $$dt=-2xdx$$ and our integral will be $$\int \frac{1}{\sqrt{t}}dt$$

Answer 2

$$\int \frac { -2x }{ \sqrt { 1-x^{ 2 } } } dx=\int \frac { d\left( 1-x^{ 2 } \right) }{ \sqrt { 1-x^{ 2 } } } =2\sqrt { 1-x^{ 2 } } +C$$

Answer 3

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:

$$\int \frac{P(x)}{\sqrt{ax^2+bx+c}} dx = Q(x)\sqrt{ax^2+bx+c} + \int \frac K {\sqrt{ax^2+bx+c}}dx$$

Answer 4

$$\int \frac{-2x}{\sqrt{1-x^2}} dx$$

$$x=\sin \theta => dx= \cos \theta\,d\theta$$

$$\int \frac{-2\sin \theta}{\sqrt{1-\sin^2\theta}} \cos \theta d\theta$$

$$\int {-2\sin \theta}{} d\theta=2\cos\theta \, +C$$ $$\arcsin x=\arcsin\sin\theta=\theta$$ $$2\cos(\arcsin x) \, +C=2\sqrt{1-x^2}+C$$

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$$\cos^2 θ + \sin^2 θ = 1\,\,,\, θ = \arcsin(x) $$

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