Homework Help: Integration help

1. Oct 9, 2004

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 ) :

Code (Text):
{  (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...

Code (Text):

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

2. Oct 9, 2004

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..

3. Oct 9, 2004

FancyNut

Thanks. :)

I've never learned those identities.. =\

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

4. Oct 9, 2004

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 $$\sin^{3}(x)$$

5. Oct 9, 2004

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*

6. Oct 10, 2004

arildno

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

7. Oct 10, 2004