Integrating -2x Over (1-x^2)^(1/2): A Confusing Challenge

[Jacobpm64]

Find the antiderivative of:

-2x ​

(1-x^2)^(1/2)

That's -2x over all of that... Bleh, i suck at code.

But anyway.. I just started integrals, and this is confusing for me... Is there a product or quotient rule in integrals like there is in derivatives?.. if not? how do you work this?

 

[d_leet]

Make a substitution, let u = 1-x² then du = -2x and you have a fairly simple integral in terms of u.

 

[dav2008]

Here is the code for it in case you need to use it in the future (click to see code)

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

 

[Jacobpm64]

how do you know what to make u and what to make du?

 

[HallsofIvy]

By thinking and analyzing the problem! The difficulty is that $\sqrt{1- x^2}$ in the denominator. The very first thing you should have thought about was substituting for that: u= 1- x^2. Then you would immediately see that du= -2xdx and that would work only if you had an xdx already in the problem (the "-2" is a constant and you can move constants in and out of the integral- you can't do that with "x", the x has to already be there) and, indeed, the problem has an "xdx". Lucky you!

(Once you've decided what to try for u, you don't "decide" what du is- that follows from the derivative of u. Notice I said "try"- you can seldom look at an integral and know what substitution will work. A lot of it is trial and error.)