how do I integrate x^3(sin5x)

Question

tkhunny · Answer

Parts, multiple times.

$\int x^{3}\sin(5x)\;dx = \int x^{3}\;d\left[-\dfrac{1}{5}\cos(5x)\right]$

Keep going!

kirbykirby · Answer

I think it's worth noting about "tabular integration by parts" which summarizes multiple integration by parts more rapidly: http://www.kkuniyuk.com/Math151TurboParts.pdf

akashdeepdeb · Answer

Try integrating by parts, Twice!

Integration by parts says:
$$\int u.dv = uv - \int v.du$$

So, to solve this question:

$$\int x^3~sin~5x~dx = -x^3~cos~ 5x + \int 3x^2~cos~5x~dx$$
You get that ^ when you set 
$u = x^3 $ and 
$dv = sin ~5x~dx$

NOW

We use the second term on the right hand side. ie. $\int 3x^2~cos~5x~dx$
Now integrate this by parts by setting
$u = cos ~5x$
$dv = 3x^2.dx$

Calculate this integral, and then substitute it in your original formula. You should get the answer.
Did you get this? :)

anonymous · Answer

uhhh none of these answers are really making sense to me, is there a table equation im supposed to be using or something?

tkhunny · Answer

Have you done any integration by parts?

anonymous · Answer

omg its been a while but i just remembered the formula you have to plug it into. haha lemme try again

tkhunny · Answer

While you are catching up, you may wish to review the material provided by kirbykirby.  It requires Way - WAY less algebra than the version I presented.  And my version is FAR less algebra than the crazy, laborious, American way AkashdeepDeb suggested.

kirbykirby · Answer

^Hehe true. I only discovered integration by parts by myself though sort of randomly, after I had done all my calculus courses . What a drag -_-

kirbykirby · Answer

^well the tabular method is what I meant hehe!

akashdeepdeb · Answer

Wow! That method is great! Thanks. :)

anonymous · Answer

wow that method seriously is great, thank you guys so much!