Integration problem using substitution

[chwala]

Homework Statement

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

Homework Equations

The Attempt at a Solution

$u= sin 4x$ [/B]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

 

[haruspex]

$u= sin 4x$ on integration $u^2/2=-cos4x/4$ ,

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

 

[chwala]

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

so should i use the second approach?

 

[haruspex]

so should i use the second approach?

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.

 

[chwala]

yes the limits are from 0 to π/24

 

[chwala]

is $0.25∫{cos^24x}du$ correct?

 

[haruspex]

the limits are from 0 to π/24

So what are the limits on u?

is $0.25∫{cos^24x}du$ correct?

Yes.

 

[chwala]

So what are the limits on u?

Yes.

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

 

[chwala]

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

should we substitute again? or are we going to have$0.25∫cos^24x d {sin4x}$

 

[chwala]

So what are the limits on u?

Yes.

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

 

[haruspex]

limits on u are 0 to 30,

No.

how do i proceed with the integration?

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?

 

[chwala]

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?

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

 

[chwala]

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?

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

 

[Eclair_de_XII]

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

 

[chwala]

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

i have seen the obstacle with that move...

 

[chwala]

i have seen the obstacle with that move...

i have seen it, check post 13

 

[SammyS]

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

i have seen the obstacle with that move...

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