# Question about trigonometric substitution

#### skatenerd

##### Active member
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

$$x={sin{\theta}}$$

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

$$\frac{\theta}{2}-\frac{sin{2\theta}}{4}$$

I know the first term is just

$$\frac{sin^{-1}x}{2}$$

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!

#### MarkFL

Staff member
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:

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

#### skatenerd

##### Active member
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