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Question about trigonometric substitution


Active member
Oct 3, 2012
Studying for finals here...So I have this specific problem to use trig substitution on.

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

I begin by substituting


I am fine with doing everything up to the point where I have an answer for the integral in terms of \(\theta\). This answer is


I know the first term is just


However the second term is always the part that throws me off. How do you find what to plug back in for \(\theta\) when the \(\theta\) is inside of a sine? Any help is appreciated!


Staff member
Feb 24, 2012
For the second term, you would use:

$\displaystyle \theta=\sin^{-1}(x)$ to get:

$\displaystyle \frac{\sin(2\sin^{-1}(x))}{4}$

Now, using the double-angle identity for sine, we have:

$\displaystyle \frac{2\sin(\sin^{-1}(x))\cos(\sin^{-1}(x))}{4}=$

Draw a right triangle here where the side opposite the angle is x, and the hypotenuse is 1, then use the Pythagorean theorem to determine the adjacent side is $\displaystyle \sqrt{1-x^2}$ and then take the cosine of this angle. We now have:

$\displaystyle \frac{x\sqrt{1-x^2}}{2}$

and so we may conclude:

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


Active member
Oct 3, 2012
Ahh wow thank you! For some reason I think I felt like a substitution inside of the trig function wasn't allowed, but now I see and that makes perfect sense. Thanks again :D