finding value of integral $$\int_0^{\pi/2} (\cos x)^{2011} \sin (2013 x) dx $$
The value is given to be $\frac{1}{2012}$. I am while solving using by parts, but one term is remaining like $\frac{1}{2}$, then $\frac{1}{3},$ and so on. Kindly help.
what i am stuck at is following.
$$I(m,n) =\int_0^{\pi/2} (\cos x)^{m} \sin (n x) dx $$
$$I(m,n)= \frac {1}{m+n} + \frac{m}{m+n}I(m-1,n-1)$$
after deriving this i am not able to proceed further
 A: Hint:
$n=m+2\implies$ $$I_{(m,m+2)}=\dfrac1{2m+2}+\dfrac m{2m+2}I_{(m-1,m+1)}$$
$I_{(0,2)}=1$
$m=1\implies$ $$I_{(1,3)}=\dfrac14+\dfrac14I_{(0,2)}=\dfrac12$$
$m=2\implies$ $$I_{(2,4)}=\dfrac16+\dfrac13I_{(1,3)}=\dfrac13$$
Use induction for $I_{(m-1,m+1)}=\dfrac2{m-1+m+1}$
A: Note that
$$
\begin{align}
I(m,n)
&=\int_0^{\pi/2}\cos^m(x)\sin(nx)\,\mathrm{d}x\tag1\\
&=\int_0^{\pi/2}\cos^{m+1}(x)\sin((n-1)x)\,\mathrm{d}x+\int_0^{\pi/2}\cos^m(x)\cos((n-1)x)\sin(x)\,\mathrm{d}x\tag2\\
&=I(m+1,n-1)-\frac1{m+1}\int_0^{\pi/2}\cos((n-1)x)\,\mathrm{d}\cos^{m+1}(x)\tag3\\
&=I(m+1,n-1)+\frac1{m+1}-\frac{n-1}{m+1}\int_0^{\pi/2}\cos^{m+1}(x)\sin((n-1)x)\,\mathrm{d}x\tag4\\
&=\frac{m-n+2}{m+1}\,I(m+1,n-1)+\frac1{m+1}\tag5
\end{align}
$$
Explanation:
$(1)$: define $I(m,n)$
$(2)$: $\sin(nx)=\sin((n-1)x)\cos(x)+\cos((n-1)x)\sin(x)$
$(3)$: prepare to integrate by parts
$(4)$: integrate by parts
$(5)$: simplify
Plug in $m=2011$ and $n=2013$ and see what happens.
However, this gives the answer $\frac1{2012}$, which is confirmed by Mathematica.
