What is the integral of (xsinx)^2?

[emin]

hi all,
I've been trying to integrate this thing for ages.i tried using integration by parts using u=x^2,
dv/dx=sin^2 x but it just doesn't seem to end.any help or pointers much appreciated.

thanks

 

[VeeEight]

Switch your u=.. and dv=.. around.
Another method would be to try an identity, such as the double angle identity for your (sinx)^2

 

[emin]

Switch your u=.. and dv=.. around.
Another method would be to try an identity, such as the double angle identity for your (sinx)^2

hmm i hadn't thought of the identity, i'll give it a go.
thanks.

 

[n!kofeyn]

I would use your original u-substitution (u=x²), and then use the half-angle formula for sin²x to integrate the dv. Then after you complete the integration by parts the first time, you'll get a sum of two functions in the integral term. One of them you'll be able to immediately integrate, while for the other one you can use integration by parts again.

 

[emin]

i managed to integrate it but i think i may ave made an error.
heres the result:
x^3/4-x^2/2sin2x-x/2cos2x-sin2x/4

don't know how to put it in proper formulae.
anyway thanks for the help.

 

[TheoMcCloskey]

emin - I come up with a slightly different answer. Can you provide details of your solution. The starting approach I took is as follows:

<br /> I = \int x^2 \cdot \sin^2(x)\,dx = \int x^2 \cdot \frac{(1-\cos(2x))}{2} \, dx<br />

then let

<br /> 2I = \int x^2 \cdot (1-\cos(2x)) \, dx = \frac{x^3}{3} - I_2<br />

where
<br /> I_2 = \int x^2 \cdot \cos(2x) \, dx = \int \left (\frac{2x}{2} \right )^2 \cdot \cos(2x) \, d\left ( \frac{2x}{2} \right ) = (1/8) \int s^2 \cdot \cos(s) \, ds<br />

where s=2x

 

[emin]

i did get my fractions mixed didn't i?
i was doing it on an a4 page and got all the working jumbled.
yes that is the approach i took, i should be able to run it through mathematica when i get the chance, and see what answer it comes up with.