If $\displaystyle J_{m, n}=\int_{0}^{\pi/2}\cos^mx\cos nx\;dx$, where $m$ and $n$ are natural numbers, then evaluate $\dfrac{J_{6,3}}{J_{5,2}}$.
My try: $$J_{m,n} = \int^{\pi/2}_{0}\cos^m x \cos nx dx = \frac {\cos^m x \sin nx}{m + n}\bigg|^{\pi/2}_{0} + \frac {m}{m + n}J_{m - 1, n - 1}$$
So we have $$J_{m,n}=\frac{m}{m+n}J_{m-1,n-1}$$
Could some Help me to solve it Thanks
The approach I tried uses two recurrences which might be easier to prove:
$$
\begin{align}
J_{m,n}
&=\int_0^{\pi/2}\cos^m(x)\cos(nx)\,\mathrm{d}x\tag1\\
&=\int_0^{\pi/2}\cos^{m-2}(x)\cos(nx)\,\mathrm{d}x-\int_0^{\pi/2}\cos^{m-2}(x)\cos(nx)\sin^2(x)\,\mathrm{d}x\tag2\\
&=J_{m-2,n}+\frac1{m-1}\int_0^{\pi/2}\cos(nx)\sin(x)\,\mathrm{d}\cos^{m-1}(x)\tag3\\
&=J_{m-2,n}-\frac1{m-1}\int_0^{\pi/2}\cos^{m-1}(x)(\cos(nx)\cos(x)-n\sin(nx)\sin(x))\,\mathrm{d}x\tag4\\
&=J_{m-2,n}-\frac1{m-1}J_{m,n}-\frac{n}{m(m-1)}\int_0^{\pi/2}\sin(nx)\,\mathrm{d}\cos^m(x)\tag5\\
&=J_{m-2,n}-\frac1{m-1}J_{m,n}+\frac{n^2}{m(m-1)}J_{m,n}\tag6\\[3pt]
&=\frac{m(m-1)}{m^2-n^2}J_{m-2,n}\tag7
\end{align}
$$
Explanation:
$(2)$: $\cos^2(x)=1-\sin^2(x)$
$(3)$: prepare to integrate by parts
$(4)$: integrate by parts
$(5)$: prepare to integrate by parts
$(6)$: integrate by parts
$(7)$: add $\frac{m(m-1)}{m^2-n^2}$ times $(6)$ to $\frac{m-n^2}{m^2-n^2}$ times $(1)$
$$
\begin{align}
J_{m-1,n+1}
&=\int_0^{\pi/2}\cos^{m-1}(x)\cos((n+1)x)\,\mathrm{d}x\tag8\\
&=\int_0^{\pi/2}\cos^{m-1}(x)(\cos(nx)\cos(x)-\sin(nx)\sin(x))\,\mathrm{d}x\tag9\\
&=J_{m,n}+\frac1m\int_0^{\pi/2}\sin(nx)\,\mathrm{d}\cos^m(x)\tag{10}\\
&=J_{m,n}-\frac nm\int_0^{\pi/2}\cos^m(x)\cos(nx)\,\mathrm{d}x\tag{11}\\[6pt]
&=\frac{m-n}{m}J_{m,n}\tag{12}
\end{align}
$$
Explanation:
$\phantom{1}(9)$: formula for the cosine of a sum
$(10)$: prepare to integrate by parts
$(11)$: integrate by parts
$(12)$: add $\frac{m-n}{m}$ times $(11)$ to $\frac nm$ times $(8)$
Applying $(7)$ and $(12)$ yields $$ \begin{align} J_{m,n} &=\frac{m(m-1)}{m^2-n^2}J_{m-2,n}\\ &=\frac{m(m-1)}{m^2-n^2}\frac{m-n}{m-1}J_{m-1,n-1}\\ &=\frac{m}{m+n}J_{m-1,n-1} \end{align} $$
Just pointing out the obvious: If $$J_{m,n}=\frac{m}{m+n} J_{m-1,n-1},$$ then we can put $m=6,n=3$ to obtain $$ \frac{J_{6,3}}{J_{5,2}} = \frac{6}{6+3} = \frac{2}{3}. $$ (However I was not able to do the first calculation)
