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I stuck into trouble when trying to solve the integral $$ \int_0^\pi\sin^{n}\theta\cos(a \cos\theta)\mathrm d\theta $$ where n is a positive integer, and $a>0$ is real number.

Using Mathematica I found that when $n$ is odd the integral is easy to do but I couldn't find a general form, and when $n$ is even the integral would contain Bessel functions $\mathrm J_1(a),\mathrm J_2(a)$

How can I handle this? Does there exist a way to find a general form of the integral?

PyroTechnics
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    When $n$ is odd, the integral is $-a^{-1}\int_0^\pi(1-\cos^2\theta)^{(n-1)/2}\cos(a\cos\theta)d(a\cos\theta)=-a^{-1}\int_a^{-a}(1-a^{-2}x^2)^{(n-1)/2}\cos(x)dx$, which can be calculated by repeatedly integrating by parts. – Kenta S Oct 05 '22 at 13:22

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Define $$F_n(a)=a^n\int_0^\pi\sin^{2n}x\cos(a\cos x)\,dx$$ Then differentiating on $a$ yields $$F_n'(a)=na^{n-1}\int_0^\pi\sin^{2n}x\cos(a\cos x)\,dx - a^n\int_0^\pi\sin^{2n}x\cos x\sin(a\cos x)\,dx$$ $$F_n''(a)=n(n-1)a^{n-2}\int_0^\pi\sin^{2n}x\cos(a\cos x)\,dx - 2na^{n-1}\int_0^\pi\sin^{2n}x\cos x\sin(a\cos x)\,dx - a^n\int_0^\pi\sin^{2n}x\cos^2x\cos(a\cos x)\,dx$$ Then we can verify that $F$ satisfies Bessel's differential equation: $$a^2F_n''(a)+aF_n'(a)+(a^2-n^2)F_n(a)=a^{n+1}\int_0^\pi\sin^{2n}x(a\sin^2x\cos(a\cos x)-(2n+1)\cos x\sin(a\cos x))\,dx$$ $$=a^{n+1}[-\sin^{2n+1}x\sin(a\cos x)]_0^\pi=0$$ Since $F_n(0)$ is well-defined for $n\ge0$, $Y_n(a)$ cannot appear in the solution and $$F_n(a)=c(n)J_n(a)$$ $c(n)$ can be determined from the series expansion of the integral, and finally we have $$\int_0^\pi\sin^nx\cos(a\cos x)\,dx=\sqrt\pi\left(\frac2a\right)^{n/2}\Gamma\left(\frac{n+1}2\right)J_{n/2}(a)$$ which holds for all nonnegative real $n$, not just integers. For odd $n$ the $J_{n/2}(a)$ term can be rewritten as its spherical counterpart $j_{(n-1)/2}(a)$, which in turn is a rational function of $\cos a,\sin a$ and $a$.

Parcly Taxel
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