# Trigonometric Substitution Problem

1. Jun 3, 2010

### Lancelot59

This problem looks relatively simple, but the coefficient in front of the variable is causing issues:

$$\int{\sqrt{1-4x^{2}}}dx$$

So I started like this:

$$x=sin(\theta)$$
$$dx=cos(\theta)d\theta$$
$$\int{\sqrt{1-4sin^{2}(\theta)}cos(\theta)d\theta}$$

Normally you can remove the constant from the root and go about the rest of the integral, but I'm stumped on this. How can I break this down into something more manageable? I've tried several identities but nothing has worked yet.

A pointer in the right direction would be great. Thanks in advance!

2. Jun 3, 2010

### Staff: Mentor

Use the substitution sin(theta) = 2x. Rather than trying to remember which substitution to use, I draw a right triangle, labellings its sides according to the quantities in the radical. My triangle has a hypotenuse of 1, opposite side of 2x, and adjacent side of sqrt(1 - 4x^2).

3. Jun 3, 2010

### Lancelot59

So you just treat the root as the Pythagorean theorem? Makes sense.

To make sure I understand this fully, if my equation was $$\int{\sqrt{1+16x^{2}}}dx$$

I would say that one of triangle's sides was 1, and the other was 4x? Making the hypotenuse the square root of 1-16x^2?

Okay, so going back to the original I can say that sin(theta) = 2x/1, and then isolate x to get x=sin(theta)/2. Putting that in it squares and cancels out the 4. Wonderful, thanks for the help!

4. Jun 3, 2010

### Staff: Mentor

Yes.
No, the hypotenuse would be sqrt(1 + 16x^2).

5. Jun 3, 2010

### Lancelot59

I hit the minus sign by accident. :p

6. Jun 3, 2010

### Staff: Mentor

I hope you don't get a job at the bank I go to :yuck:

(Just kidding!)

7. Jun 4, 2010

### Lancelot59

Eww...banking.:yuck: