Expert: Paul Klarreich - 2/16/2008

Question
I am asked to integrate

[sin^3(x) cos^3(x)]

SO:

u=sin(x)
du=cos(x) dx

Now:

sin^3(x) cos^2(x) cos(x) dx
(sin^3(x)) (1- sin^2(x)) (cos(x) dx)

u^3 - u^5 du

Leads to:

(1/4)(sin^4(x)) - (1/6)(sin^6(x)) + C


When I take the derivative of this (as a check)
I come back to  [sin^3(x) cos^3(x)]

My book has

(1/48)(cos^3(2x)) - (1/16)(cos(2x))

Are we both correct. (Maybe my book wants me to use
a different method to solve?)  

Answer
Questioner:   Mark
Category:  Calculus
Private:  No
 
Subject:  Integration
Question:  I am asked to integrate

[sin^3(x) cos^3(x)]

SO:

u=sin(x)
du=cos(x) dx

Now:

sin^3(x) cos^2(x) cos(x) dx
(sin^3(x)) (1- sin^2(x)) (cos(x) dx)

u^3 - u^5 du

Leads to:

(1/4)(sin^4(x)) - (1/6)(sin^6(x)) + C


When I take the derivative of this (as a check)
I come back to  [sin^3(x) cos^3(x)]

My book has

(1/48)(cos^3(2x)) - (1/16)(cos(2x))

Are we both correct. (Maybe my book wants me to use
a different method to solve?)
....................................
Hi, Mark,

One of the funny things about studying Calculus is in finding that the book's answer does not appear to be the same as yours, and trying to reconcile the two. [I know what you are thinking -- this guy has a warped sense of humor.]

BUT:  
{
|[sin^3(x) cos^3(x)]
}

Use the fact that  sin x cos x = (1/2) sin(2x), and you have:

{
|1/8 sin^3(2x) dx
}

{
|1/8 sin(2x)(1 - cos^2(2x)) dx
}

Now u = cos(2x),  du = - 2 sin(2x) dx

{
|- 1/16 (1 - u^2) du
}

- 1/16 (u - u^3/3)

- cos(2x)/16 + cos^3(2x)/48

Yes, that's the book's answer.  So you are right in this respect -- the book [that means the grad student who got paid a few pennies to write the answers] expects you to use a different method.

Now what about reconciling?

Use a dbl-angle formula:

cos(2x) = 1 - 2 sin^2(2x)

Now the book's answer:

- cos(2x)/16 + cos^3(2x)/48 =

[writing s = sin x to save typing.]

- (1 - 2s^2)/16 + (1 - 2s^2)^3/ 48  << dbl angle formula used.

- (1 - 2s^2)/16 + (1 - 6s^2 + 12s^4 - 8s^6)/ 48  << binomial expansion

- 1/16 + s^2/8 +  1/48 - s^2/8 + s^4/4 - s^6/6

- 1/16  +  1/48  + s^4/4 - s^6/6

Now that is, indeed, your answer.  [Forget about the -1/16 and +1/48, they get subsumed into the +C you are supposed to write.]

So, as you find in http://math.boisestate.edu/gas/mikado/libretto.txt:

                   And I am right,
                   And you are right,
              And everything is quite correct!