Integration of (xsinx)^2

1. Oct 2, 2004

josephcollins

Hi ppl, I need a little pointer with the integral of (xsinx)^2. I tried by parts but it just doesn't stop and I've tried writing sin^2x in it's other forms but that yields similar results. Any help please?

Joe

2. Oct 2, 2004

arildno

If it doesn't stop, it is because you switch what's "u" and "v".
It is definitely smart to use the double-angle formula!

3. Oct 2, 2004

Pyrrhus

Use

$$sin^{2}x = \frac{1-cos2x}{2}$$

4. Oct 2, 2004

josephcollins

I used the identity suggested, but I don't get anywhere, could u please just take the problem a little further?

5. Oct 2, 2004

arildno

Show in some detail why you don't get anywhere!

6. Oct 3, 2004

irony of truth

By the way, may I "join"? In the first place, do you already know the answer in the first hand? (I mean, you got the answer somewhere else? =) )

Since the members here are not giving you a direct answer to the question... I'll give you some hints... I only did some integration by parts....

1.) let u = sin^2 x and dv = x^2dx... so, du = sin2x dx and v = x^3/3...
2.) You get the 2nd integral, right? If you've done the first step (of mine) correctly, then your on the track..
let u = x^3 and dv = sin[2x] dx... so, du = 3x^2 dx and v = -cos[2x] / 2.
3.) keep doing integration by parts a little more... later, you'll notice something - integration by simple substitution... and then if there's anything to simplify, do so.

Show your work... so that others can guide you...

Cyclovenom's hint for you, unfortunately, does not apply in my strategy that I've written for you here... but still, both different approach, if correctly done, can lead you to the same answer. =) Good luck!

7. Oct 3, 2004

irony of truth

Joe, I hope that helps... I'll be watching here.. time to time. =)

8. Oct 7, 2004

josephcollins

Thanks for the help irony, I managed it now, cheers,
Joe

9. Oct 8, 2004

irony of truth

Welcome... :D