Evaluate $\int_0^{\pi/2} \sqrt{ \cos x}\>\cos\frac{3x}2 \>\cos(2n+1)x\> dx$ I am interested in the trigonometric integral below
$$I_n = \int_0^{\pi/2} \sqrt{
\cos x}\>\cos\frac{3x}2 \>\cos(2n+1)x\> dx
$$
A direct integration appears a bit too much. With the help of an online integration tool for a number of $n$’s, I was able to guess the recursive relationship below
$$I_n=-\frac{2n-3}{2n}I_{n-1}$$
which I verified numerically as well. So, the recursion looks like a promising approach, except I was still not able to derive it analytically.
 A: There is a "beautiful" (bad joke !) antiderivative in terms of Guassian hypergeometric functions. Using the beta function, it write
$$J_n=i\frac{ (-1)^n \sqrt{\cos (x)}}{8 \sqrt{1+e^{2 i x}}}\,(T_1-T_2)$$
$$T_1=e^{-\frac{i x}{2}} \sqrt{-e^{2 i x}} \left(B_{-e^{2 i
   x}}\left(-n-\frac{3}{2},\frac{3}{2}\right)-B_{-e^{2 i
   x}}\left(n-\frac{1}{2},\frac{3}{2}\right)\right)$$
$$T_2=e^{\frac{i x}{2}} \left(B_{-e^{2 i x}}\left(-n,\frac{3}{2}\right)+B_{-e^{2 i
   x}}\left(n+1,\frac{3}{2}\right)\right)$$
Now computing
$$I_n=(-1)^{n+1}\,\frac{1}{8}\,\sqrt{\frac{\pi }{2}}\,\, \frac{\Gamma \left(n-\frac{1}{2}\right)}{\Gamma (n+1)}$$
A: I shall provide some hints on how to proceed.
Rewrite your recursive formula as follows,
$$\frac{I_{n}}{I_{n-1}}=\frac{3-2n}{2n}$$.
Again apply the recursion,
$$\frac{I_{n-1}}{I_{n-2}}=\frac{3-2(n-1)}{2(n-1)}$$...
Repeat this process until the least possible convergent value of $n$ and multiply all of them. If you notice all of them cancels off leaving $I_{n}$ which is required. The least possible convergent value is also left but can be easily evaluated.
