Integrating squares of sine & cosine

[sg001]

Homework Statement

find ∫4cosx*sin^2 x.dx

Homework Equations

The Attempt at a Solution

∫4cos x * 1/2 (1 - cos2x)

∫2cosx - 4cos^2 x.

Then i don't know whereto go from here??

 

[kushan]

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

 

[Mark44]

Homework Statement

find ∫4cosx*sin^2 x.dx

Homework Equations

The Attempt at a Solution

∫4cos x * 1/2 (1 - cos2x)

∫2cosx - 4cos^2 x.

Then i don't know whereto go from here??

A simple substitution will work in this problem.

 

[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.

 

[sg001]

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

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

= 2cosx - 4cos^2 x

 

[Mark44]

ie,
let u = cos x

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.

∫ 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.

 

[SammyS]

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.

Instead:

Let u = sin(x) .

 

[Mark44]

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

= 2cosx - 4cos^2 x

No.
cos(x) * cos(2x) \neq cos²(x)

 

[sg001]

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

 

[kushan]

Yea I asked that too

 

[sg001]

∫4cosx * u^2 * du/cosx

∫4(u)^2

4/3 (u)^3 + C

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

Better?

 

[kushan]

Now you can proceed , easily :)

 

[Mark44]

∫4cosx * u^2 * du/cosx

∫4(u)^2

4/3 (u)^3 + C

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

Better?

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

∫4cosx * u^2 * du/cosx

=[/color]∫4(u)^2 du[/color]

=[/color]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

 

[sg001]

okie