I know that we already have some symmetrical looking reduction formulas for similar functions like:
$$I_{m,n} = \int \cos^m(x) \sin(nx) dx = -\frac{\cos^m(x)\cos(nx)}{m+n} + \frac{m}{m+n} I_{m-1,n-1}$$
$$I_{m,n} = \int \cos^m(x) \cos(nx) dx = \frac{\cos^m(x)\sin(nx)}{m+n} + \frac{m}{m+n} I_{m-1,n-1}$$
So, I was expecting something similar for $\int \sin^m(x)\sin(nx)dx$, however simply doing integration by parts as is used in above two didn't worked for it.
I am a bit new to integration, So pardon me if it is something silly. Any help would be greatly appreciated.
Best Answer
Disclaimer: I got a recurrence along $m$, but it is somewhat of a complicated expression. Specifically, it is $$I_{m,n} = \frac{m(m-1)I_{m-2,n}- m\sin(nx)\cos(x)\sin^{m-1}(x)+n\cos(nx)\sin^m(x)}{m^2-n^2}$$
$$I_{m,n} = \int \sin^m(x) \sin(nx)dx$$ Factor out $\sin^2(x) = 1-\cos^2(x)$ to make it $$I_{m,n} = \int \sin^{m-2}(x) \sin(nx)dx-\int\sin^{m-2}(x) \sin(nx)\cos^2(x) dx$$
Or equivalently $$I_{m-2, n} - \int\sin^{m-2}(x) \sin(nx)\cos^2(x) dx$$
Then using integration by parts makes it $$I_{m-2, n} - \frac{\sin(nx)\cos(x)\sin^{m-1}(x)}{m-1}+\frac{1}{m-1}\int\left( n\cos(x)\cos(nx)\sin^{m-1}(x) - \sin(nx)\sin^{m}(x) \right) dx$$
Since that last part is equal to the original integral, $$I_{m,n}=\frac{m-1}{m}I_{m-2,n}- \frac{\sin(nx)\cos(x)\sin^{m-1}(x)}{m}+\frac{n}{m}\int\left(\cos(x)\cos(nx)\sin^{m-1}(x) \right) dx$$
Using integration by parts again yields $$\frac{m-1}{m}I_{m-2,n}- \frac{\sin(nx)\cos(x)\sin^{m-1}(x)}{m}+\frac{n}{m}\left(\frac{\cos(nx)\sin^m(x)}{m}+\frac{n}{m}I_{m,n}\right)$$
This then means that $$I_{m,n} = \frac{m(m-1)I_{m-2,n}- m\sin(nx)\cos(x)\sin^{m-1}(x)+n\cos(nx)\sin^m(x)}{m^2-n^2}$$
which is a recurrence along $m$.