# Homework Help: Integration problem using substitution

1. Feb 1, 2017

### chwala

1. The problem statement, all variables and given/known data

using $u= sin 4x$ find the exact value of $∫ (cos^3 4x) dx$

2. Relevant equations

3. The attempt at a solution

$u= sin 4x$
on integration $u^2/2=-cos4x/4$ , →$-2u^6={cos 4x}^3$......am i on the right track because now i end up with $∫{{-2u^6}/{4.-2u^2}}du$ or should i use

$du=4cos 4x dx$ to end up with $0.25 ∫ cos^24x du$ which looks wrong to me

2. Feb 1, 2017

### haruspex

No, you've integrated one side wrt u and the other wrt x.

3. Feb 1, 2017

### chwala

so should i use the second approach?

4. Feb 1, 2017

### haruspex

Yes, but you need to get all of the references to x turned into references to u.

You are asked for an exact value, but it is an indefinite integral. Remember that the limits need to be expressed in terms of u as well.

5. Feb 1, 2017

### chwala

yes the limits are from 0 to π/24

6. Feb 1, 2017

### chwala

is $0.25∫{cos^24x}du$ correct?

7. Feb 1, 2017

### haruspex

So what are the limits on u?
Yes.

8. Feb 1, 2017

### chwala

limits on u are 0 to 30, now how do i proceed with the integration?

9. Feb 1, 2017

### chwala

$cos^2 4x$ = $(cos 8x+1)$/2

should we substitute again? or are we going to have

$0.25∫cos^24x d {sin4x}$

10. Feb 1, 2017

### chwala

i am a bit confused we cannot integrate a variable say $x$ with respect to another variable say $u$, i am stuck here

11. Feb 1, 2017

### haruspex

No.
You have the cos2 of some angle, and you need to express that in terms of u, the sine of the same angle. Does nothing click?

12. Feb 1, 2017

### chwala

sorry limits are from 0 to 0.5 an oversight on my part..........

13. Feb 1, 2017

### chwala

i now get it lol
$0.25∫{1-u^2}du$ from u=0 to u=0.5 thanks mate solution is $0.115$

14. Feb 1, 2017

### Eclair_de_XII

Why don't you try splitting $cos^34x$ into $cos4x$ and another term containing the term used for $u$ substitution?

15. Feb 1, 2017

### chwala

i have seen the obstacle with that move.........

16. Feb 1, 2017

### chwala

i have seen it, check post 13

17. Feb 2, 2017

### SammyS

Staff Emeritus
@chwala
Actually that is the move you finally made to solve. Check the time of @Eclair_de_XII 's post and your posts.