I have the following integration
$\int_{-\pi}^{\pi} \frac{1} {\sqrt\pi} \sin (n\frac{(x+\pi)} {2})x^2 \frac{1} {\sqrt\pi} \sin (m\frac{(x+\pi)}{2}) dx$ $m$ and $n$ are integers,
and I used Maple to get the integration and the result is $\frac{(4(4(-1)^{n+m}nm+4nm)}{m^4-2m^2n^2+n^4}$
Actually, I'm not sure about the result, I dont know if there is any other way you can suggest to compute the integration rather than the integration by parts?
Thank you in advance..
Working the antiderivative, you could start rewriting the product of the sines by the differences of cosines.
Doing it, you face two integrals looking like $$\int \cos(a x+b)\,x^2\,dx$$ Now, assume that the result is of the form $P \cos(ax+b)+Q \sin(ax+b)$ where $P$ and $Q$ are polynomials.
So, differentiating both sides $$\cos(a x+b)\,x^2=(a Q+P') \cos(ax+b)-(aP-Q')\sin(ax+b)$$ Because of the cosine term the degree cannot be larger than $2$ for $aQ+P'$ making as a maximum $Q_2$ and $P_3$.
So, let $P=\sum_{i=0}^3 \alpha(i) x^i$ and $Q=\sum_{i=0}^2 \beta(i) x^i$ and replace.
The rhs will become $$[(a \beta (0)+\alpha (1))+x (a \beta (1)+2 \alpha (2))+x^2 (a \beta (2)+3 \alpha (3))] \cos(a x+b)+$$ $$[(a \alpha (0)-\beta (1))+x (a \alpha (1)-2 \beta (2))+a \alpha (2) x^2+a \alpha (3) x^3] \sin(a x+b)$$ Now, identify to get the value of the coefficients and get $$\left( \begin{array}{ccc} i & \alpha_i &\beta_i \\ 0 & 0 & -\frac{2}{a^3} \\ 1 & \frac{2}{a^2} & 0 \\ 2 & 0 & \frac{1}{a} \\ 3 & 0 & \end{array} \right)$$ making $$\int \cos(a x+b)\,x^2\,dx=\frac{2 a x \cos (a x+b)+\left(a^2 x^2-2\right) \sin (a x+b)}{a^3}$$