Calculus 2 question (∫cos^3(x)dx)

[aglo6509]

Homework Statement

∫cos^3(x)dx

Homework Equations

http://college.cengage.com/mathemat...students/derivatives/derivative_integrals.pdf

http://college.cengage.com/mathemat...ytic/8e/students/trig_review/trigonometry.pdf

This is the website I've using to get my formulas.

The Attempt at a Solution

This seems to be an easy problem but I can't the answer.

Heres my attempt:

∫cos^3(x)dx
∫cosxcos^2(x)dx
∫cosx((1+cos2x)/2) dx
∫(cosx + (cosxcos2x))/2 dx
∫(cosx)/2dx + ∫(cosxcos2x)/2 dx
-1/2sinx + ∫1/2(cos(-x)+cos3x)dx <---double angle formula
-1/2sinx + 1/2[sin(-x) - 1/3sin3x] +C

I thought this was right, but the answer they give in the back of my exam review packet is:

sinx-(sin^3(x))/3 +C

I can't think of anyway of getting a sin^3! Please help

 

[micromass]

Hi aglo6509,

Heres my attempt:

∫cos^3(x)dx
∫cosxcos^2(x)dx
∫cosx((1+cos2x)/2) dx
∫(cosx + (cosxcos2x))/2 dx
∫(cosx)/2dx + ∫(cosxcos2x)/2 dx
-1/2sinx + ∫1/2(cos(-x)+cos3x)dx <---double angle formula

Firstly, $\int{\cos(x)dx}=\sin(x)+C$ and not -sin(x), like you wrote.
Secondly, I really don't see how you used the "double angle formula" there. In fact, whatever formula you used, you probably made a mistake since the transition is not correct. That is $\cos{x}cos(2x)\neq \cos(-x)+\cos(3x)$.

-1/2sinx + 1/2[sin(-x) - 1/3sin3x] +C

I thought this was right, but the answer they give in the back of my exam review packet is:

sinx-(sin^3(x))/3 +C

I can't think of anyway of getting a sin^3! Please help

Also note, that we have $\sin(3x)=3\sin(x)-4\sin^3(x)$. So, this is how the cube comes into the problem.

Furthermore, you've chosen a quite difficult approach to the problem. Did you already see substitution? In that case, if you substitute t=sin(x) in the integral [itex]\int{\cos^3(x)dx}[/tex], then the integral becomes much easier![/itex]

 

[JThompson]

This problem is much easier if you use the formula cos^2(x) = 1-sin^2(x). In general, try simple formulas before more complicated ones. (This is of course relative, but I classify a simple formula as one that comes up often and is extremely easy to remember.)

Also ∫(cosx)/2dx is 1/2sinx, not -1/2sinx.

 

[aglo6509]

Oh I meant Product-to-Sum formula not a double angle formula.

 

[micromass]

Oh I meant Product-to-Sum formula not a double angle formula.

I thought so
But you still made a mistake in applying the formula. You forgot a factor 1/2 in front of the sum. (Remember that you already had a factor 1/2, so together you should have a factor 1/4)

 

[aglo6509]

Wow guys I just realized the big thing I looked over.

So it would be:

∫cos^3(x)dx
∫cosxcos^2(x)dx
∫cosx(1-sin^2(x))dx
∫cosx-sin^2(x)cosxdx
∫cosxdx-∫sin^2(x)cosxdx
sinx - (sin^3(x))/3 + C

Thanks guys, I looked at it again and within 3 seconds saw my error

 

[Ray Vickson]

If you change variables to y = sin(x), so dy = cos(x) dx, you have integral (1 - y^2) dy, which looks pretty easy.

RGV