Integrating Problems: Solving a Cubic Integral

[FancyNut]

Hi guys... I've been working on this problem for 30 minutes... yes I'm stupid. Seriously I'm thinking right now what the hell I'm going to do. Out of every 4 problems I get 3 wrong and re-do them... I understand where I went wrong but I'm not getting any better. Scary times.

Well here it is (btw { = integral of ) :

{  (x)^3 * (cos (x))^3 dx

the second part is cosine cubed (not x).

Here's what I got after doing integration by parts 3 times...

x^3 * (sin(x))^2 + x^2 * (cos(x))^2 - 2 * (sin(x))^3

 

[arildno]

1. You ought to rewrite your cubed cosine with trigonometric identities:
\cos^{3}(x)=\cos^{2}(x)\cos(x)=\frac{1+\cos(2x)}{2}\cos(x)
\cos(x)\cos(2x)=\frac{1}{2}(cos(x)+\cos(3x))
That is,
\cos^{3}(x)=\frac{1}{4}(3\cos(x)+\cos(3x))
2. Use this expression when doing integration by parts..

 

[FancyNut]

Thanks. :)

I've never learned those identities.. =\

btw is there a method to solving it without breaking cosine cubed into smaller pieces?

 

[arildno]

Sure; if you are careful with in your calculations, you may do as follows:
\cos^{3}(x)=\cos(x)-\sin^{2}(x)\cos(x)
Note that we have the following anti-derivative:
\int\sin^{2}(x)\cos(x)dx=\frac{1}{3}\sin^{3}(x)
An analogus decomposition may then be used for \sin^{3}(x)

 

[FancyNut]

awesome. :D

I have one last question though. ^_^;;

How to solve the same integral but the cosine is not cubed... that would be:

{ x^3 * cos(x^2) dx

I also ended up doing integration by parts 3 times on this one.

*hides under desk*

 

[arildno]

\int{x}^{3}\cos(x^{2})dx=\frac{x^{2}}{2}\sin(x^{2})-\int{x\sin(x^{2})}dx
\int{x}^{3}\cos(x^{2})dx=\frac{x^{2}}{2}\sin(x^{2})+\frac{\cos(x^{2})}{2}

 

[FancyNut]

Thanks for replying. :)

So you made u = x^3? doesn't that mean du should be 3x^2?