Integration problem

1. Jan 30, 2006

zee_22

Use the identity Sin[x]^2 = [1-Cos[2x]]/2
to help calculate integral from 0 to t Sin[x]^2 dx
This question seems really easy but im having some difficulty with it.
This is what im thinking :
first put the [1-Cos[2x]]/2 instead of the Sin[x]^2
so this is what i have now (int from 0 to t) [1-Cos[2x]]/2dx
then i thought let u = cos[2x] then du = - Sin[2x]dx
and when x= 0 then u = 1 and when x= t then u= cos[2t]
then (int from 0 to t) [1-Cos[2x]]/2dx
=-1/2 (int from 1 to cos[2t] ) [1-u]/[Sin[2x]du]

2. Jan 30, 2006

d_leet

Why did you make a substitution after you used the trig identity to get rid of the sine squared, cos(2x) is certainly integrable.

3. Jan 30, 2006

zee_22

back to the integration problem

the substitution was the first thing i did i should leave the sin[x]^2 like it is then??
let sin[x]= u ???
what is the integral of Cos[2x]??

4. Jan 30, 2006

zee_22

Ok i think i got it:shy:
I just one clerification is sin^2[x] the same as Sin[x]^2???

5. Jan 30, 2006

arildno

Do NOT post homework questions or any other questions in the tutorials section! :grumpy:

Warning sent off to mods.

6. Jan 30, 2006

zee_22

im very sorry ,I'm new ,i didn't know..

7. Jan 30, 2006

Jameson

Yes that's correct. That notation is used to not confuse the following expressions.

$$\sin(x)^2$$ and $$\sin(x^2)$$

8. Jan 31, 2006

aiglet_2000

not much trouble by the way
but a slight mistake
if u=cos2x
than d(u)=-sin2xd(2x)
=-2sin2xdx
but i dont really know why u r making this substitution?
cos2x is a basic integration, but if u want to do it this way which is useless than do tell i will tell u than how to do this useless thing

9. Jan 31, 2006

aiglet_2000

u rnt using [x] to denote gretest integer function? that would make it slightly complex

10. Jan 31, 2006

Jameson

No, he's not using [x] as the greatest integer function. Anyway, here's the original integral.

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

No constant of integration because it's a definite integral. Just plug in your bounds.