For integers $a,n\in \mathbb N$, consider the following integral $$ I_n(a) = \frac{(-i)^x}{\pi}\int_0^\pi e^{i\theta(n-2a)} \sin^x \theta \cos^{n-x} \theta\; \mathrm d\theta\;. $$ How would one go around evaluating such an integral?
To be completely honest, I am being a bit cheeky here since I know the answer to the question, as I obtained this integral by applying the Cauchy integral formula on the generating function of Kravchuk polynomials. To see the answer, you may point your mouse at the answer paragraph below.
What I have been trying to find without success is a direct calculation of this integral (without using the Cauchy integral formula). I have no experience with integrals, so I am most interested in seeing what heuristic rules one takes when evaluating an integral of products of cosines and sines.
I am guessing that such products of sines and cosines is seen frequently seen in integrals and hoping there is known way to work with them.
Thank you!
The value of $I_n(a)$:
Consider the expression $\left(\frac{1-z}{2}\right)^x\left(\frac{1+z}{2}\right)^{n-x}$ as a function of $z$.
The coefficient of $z^a$ is $\sum_{j=0}^a(-1)^j\binom{x}{j}\binom{n-x}{a-j}\Big/2^n $.
Applying Cauchy integral formula on the above function, we get that the coefficient of $z^a$ is also $\frac{1}{2\pi}\oint_0^{2\pi}((1-z)/2)^x((1+z)/2)^{n-x}z^{-(a+1)}\;\mathrm d z$. Now substituting $z=e^{2i\theta}$, and integrating over the unit disk in the counter clockwise direction, we obtain that this latter integral is equivalent to $I_n(a)$.