# Thread: [SOLVED] Simple trig integration

1. ## [SOLVED] Simple trig integration

Can you guys help. I haven't done this stuff in a while.

How would you integrate:
(sin x)^2

Thanks if you can shed light.

2. Originally Posted by Evales
Can you guys help. I haven't done this stuff in a while.

How would you integrate:
(sin x)^2

Thanks if you can shed light.
Recall the double angle formula $\cos (2x) = 1 - 2 \sin^2 x$ .....

3. Originally Posted by Evales
Can you guys help. I haven't done this stuff in a while.

How would you integrate:
(sin x)^2

Thanks if you can shed light.
note that $\sin^2 x = \frac {1 - \cos 2x}2$. integrate that. it is easier

EDIT: Thanks a lot, Mr Fantastic!

4. Originally Posted by Jhevon
note that $\sin^2 x = \frac {1 - \cos 2x}2$. intgrate that. it is easier
What a dynamic duo ..... Note how posts #2 and #3 just seamlessly flow

5. Originally Posted by mr fantastic
What a dynamic duo ..... Note how posts #2 and #3 just seamlessly flow
Yup, can't help it, being around the Flow Master and all

6. Never heard of that particular identity before.

I assume that in order to do this I shouldn't use identities that aren't on the syllabus. Anyway. Getting past that. How would I do it? Should I use substitution or the chain rule ect.

7. Originally Posted by Evales
Never heard of that particular identity before.

I assume that in order to do this I shouldn't use identities that aren't on the syllabus. Anyway. Getting past that. How would I do it? Should I use substitution or the chain rule ect.
$\cos(2x)=\cos^2(x)-\sin^2(x)$
from that we can replace

$\cos^2(x)=1-\sin^2(x)$

to get

$\cos(2x)=1-2\sin^2(x)$

From there you can find what they said

so

$\int\sin^2(x)dx=\int\frac{1-\cos(2x)}{2}dx=\frac{x}{2}-\frac{1}{2}\int\cos(2x)dx$

for the last one isolate the derivative of the quanity and be done with it =D

8. Okay thanks guys.

Just wondering where how the function comes to be divided by two.

I've gotten to
(sin x)^2 = 1 - cos(2x)

9. Originally Posted by Evales
Okay thanks guys.

Just wondering where how the function comes to be divided by two.

I've gotten to
(sin x)^2 = 1 - cos(2x)
$\cos(2x)=1-2\sin^2(x)\Rightarrow{2\sin^2(x)=1-\cos(2x)}\Rightarrow{\sin^2(x)=\frac{1-\cos(2x)}{2}}$

10. Oh lol Duh.

Thanks for holding my hand through this process!

11. $\int{\sin ^{2}x\,dx}=\int{\sin x(-\cos x)'\,dx}=-\sin x\cos x+\int{\cos ^{2}x\,dx}.$

Now use $\cos^2x=1-\sin^2x$ and the conclusion follows.