Integrating sin^2x: Trig Substitution or Identity? | Quick Question"

[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.

 

[Gunthi]

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.

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

 

[Solar Eclipse]

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

 

[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?

 

[Solar Eclipse]

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

 

[Anonymous217]

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