How do you integrate #int (3x)/sqrt((1-x^2))dx# using trigonometric substitution?

Question

Isabella Ackerman · Accepted Answer

You need to use the beautiful Pythagorean right triangle.

Here is the general antiderivative :
#-3*sqrt(1- x^2)#

Here's a quick paint job to show you the lovely right triangle:

You now need to find the following:

#1)x#

#2)dx/(d theta)#

#3)sqrt(1-x^2)#

I'll give you a gentle and loving push in the right direction:

What is x?

Well, it sure looks to be :

#x = sin (theta)#

Your turn!

William Abdalla · Answer

undefined To integrate $\int \frac{3x}{\sqrt{1-x^2}} \, dx$ using trigonometric substitution, you can let $x = \sin(\theta)$, which implies $dx = \cos(\theta) \, d\theta$. Substituting these into the integral yields:

$$
\begin{align*}
\int \frac{3x}{\sqrt{1-x^2}} \, dx &= \int \frac{3\sin(\theta)}{\sqrt{1-\sin^2(\theta)}} \cos(\theta) \, d\theta \
&= \int \frac{3\sin(\theta)}{\cos(\theta)} \cos(\theta) \, d\theta \
&= \int 3\sin(\theta) \, d\theta \
&= -3\cos(\theta) + C
\end{align*}
$$

Finally, resubstitute $\theta$ in terms of $x$ to get the final result:

$$
\int \frac{3x}{\sqrt{1-x^2}} \, dx = -3\cos(\arcsin(x)) + C
$$

Alternatively, you can use the identity $\cos(\arcsin(x)) = \sqrt{1-x^2}$ to simplify the answer to $-3\sqrt{1-x^2} + C$.