# Question on an Integral

1. Feb 16, 2010

### Anonymous217

Hi guys, I have a quick question.
When integrating $$\int sin^2xdx$$, do you need to use the trig substitution $$\frac{1-cos(2x)}{2}$$?
I was able to integrate it using just $$1-cos^2x$$ as my identity and even when I showed a calc teacher how I solved the problem, he doesn't believe me or my math for some reason even though I proved that the two methods get the same result but in a different form.

2. Feb 16, 2010

### Gunthi

The best way to know is for you to post your math.

3. Feb 16, 2010

### Solar Eclipse

That looks like a textbook integration by parts problem. Could that be how your teacher wants you to solve it?

4. Feb 16, 2010

### Anonymous217

Yeah, I did it using integration by parts even though I was supposed to use the trig substitution.
$$\int sin^2xdx$$
$$Let u = sinx, dv = sinxdx$$
$$du = cosxdx, v = -cosx$$
$$-sinxcosx + \int cos^2xdx$$
$$\int sin^2xdx = -sinxcosx + \int 1 - sin^2xdx$$
$$\int sin^2xdx = -sinxcosx + x - \int sin^2xdx$$
$$2\int sin^2xdx = -sinxcosx + x$$
$$\int sin^2xdx = \dfrac{1}{2}x - \dfrac{1}{2}sinxcosx$$

Both should work right? You don't "have" to use the cos(2x) thing right?

5. Feb 16, 2010

### Solar Eclipse

They both definitely work. But whether or not you have to use Trig.Sub. is up to who ever controls your grade.

6. Feb 17, 2010

### Anonymous217

Okay cool. And this wasn't for a grade or anything just to be clear.