# Integrating squares of sine & cosine

1. Feb 23, 2012

### sg001

1. The problem statement, all variables and given/known data
find ∫4cosx*sin^2 x.dx

2. Relevant equations

3. The attempt at a solution

∫4cos x * 1/2 (1 - cos2x)

∫2cosx - 4cos^2 x.

Then i dont know whereto go from here??

2. Feb 23, 2012

### kushan

your last step how did you reach to 4cos^2 x ?

3. Feb 23, 2012

### Staff: Mentor

A simple substitution will work in this problem.

4. Feb 23, 2012

### sg001

ie,
let u = cos x

∫ 2u - 4u^2

= u^2 -4/3 (u)^3 + C

This still does not lead me anywhere?

Or did you mean a substitution not in terms of 'u' but in terms of another trig result?
I feel lost.

5. Feb 23, 2012

### sg001

by expanding 4cosx * 1/2(1-cos2x)

= 2cosx - 4cos^2 x

6. Feb 23, 2012

### Staff: Mentor

1. Start from the original integral.
2. That's almost the right substitution, but what is du? You are being very sloppy by omitting the differential, and this is exactly the type of problem where doing that will bite you in the butt.

Last edited: Feb 23, 2012
7. Feb 23, 2012

### SammyS

Staff Emeritus

Let u = sin(x) .

8. Feb 23, 2012

### Staff: Mentor

No.
cos(x) * cos(2x) $\neq$ cos2(x)

9. Feb 23, 2012

### sg001

Hmmm
I think i need to stop rushing through my questions.
Thanks for pointing out the obvious!!

10. Feb 23, 2012

### kushan

11. Feb 23, 2012

### sg001

∫4cosx * u^2 * du/cosx

∫4(u)^2

4/3 (u)^3 + C

= 4/3 (sin x)^3 +C

Better?

12. Feb 23, 2012

### kushan

Now you can proceed , easily :)

13. Feb 23, 2012

### Staff: Mentor

Yes, better, and that's the right answer, but some of the supporting work could be improved.

∫4cosx * u^2 * du/cosx

=∫4(u)^2 du

=4/3 (u)^3 + C

= 4/3 (sin x)^3 +C

In addition, it's usually better to make the substitution completely so that all the x's and dx's change to u's and du's, like this:

∫4cos(x) * sin2(x)dx [u = sin(x), du = cos(x)dx]
= 4 ∫u2 du
= 4 u3/3 + C
= (4/3) sin3(x) + C

14. Feb 23, 2012

okie