Problem source

Let ${\displaystyle I_{m,n}=\int\cos^{m}x\sin nx\,\mathrm{d}x,}$
where $m$ and $n$ are non-negative integers. Prove that for $m,n\geqslant1,$
\[
(m+n)I_{m,n}=-\cos^{m}x\cos nx+mI_{m-1,n-1}.
\]
\begin{questionparts}
\item Show that ${\displaystyle \int_{0}^{\pi}\cos^{m}x\sin nx\,\mathrm{d}x=0}$
whenever $m,n$ are both even or both odd. 
\item Evaluate ${\displaystyle \int_{0}^{\frac{\pi}{2}}\sin^{2}x\sin3x\,\mathrm{d}x.}$
\end{questionparts}

Solution source

Let ${\displaystyle I_{m,n}=\int\cos^{m}x\sin nx\,\mathrm{d}x,}$

Then
\begin{align*}
&& I_{m,n} &= \int\cos^{m}x\sin nx\,\mathrm{d}x \\
&&&= \left [ -\frac1n \cos^m x \cos n x \right] - \frac{m}{n} \int \sin^{m-1} x \cos x \cos n x \d x \\
&&&= \left [ -\frac1n \cos^m x \cos n x \right] + \frac{m}{n} \int  \sin^{m-1} x (\cos (n-1)x -\sin x \sin nx) \d x\\
&&&= \left [ -\frac1n \cos^m x \cos n x \right] + \frac{m}{n} \int  \sin^{m-1} x \cos (n-1)x \d x-\frac{m}{n} I_{m,n} \\
&&&= \left [ -\frac1n \cos^m x \cos n x \right] + \frac{m}{n} I_{m-1,n-1} -\frac{m}{n} I_{m,n} \\
\Rightarrow && nI_{m,n} &= -\cos^m x \cos n x + mI_{m-1,n-1} -mI_{m,n}\\
\Rightarrow && (m+n)I_{m,n} &=  -\cos^m x \cos n x + mI_{m-1,n-1}
\end{align*}

\begin{questionparts}
\item Note that $I_{2m,0} = 0$ (the integrand is 0) and $I_{0, 2m} = 0$ (symmetry for our limits).

$\displaystyle \left [-\cos^m x \cos n x \right]_0^\pi = \l - (-1)^m (-1)^n \r - \l -1 \r = 1 - (-1)^{m+n} = 0$ since $m+n$ is even. Therefore all reductions are $I_{m,n} = \frac{I_{m-1,n-1}}{m+n}$ terminating at $0$, so all values are zero

\item 
\begin{align*}
 \int_{0}^{\frac{\pi}{2}}\sin^{2}x\sin3x\,\mathrm{d}x &= \int_{0}^{\frac{\pi}{2}}(1-\cos^2x)\sin3x\,\mathrm{d}x \\
&= I_{0,3} - I_{2,3} \\
&= \frac13 - \frac15 \l \left [-\cos^2 x \cos 3 x \right]_0^{\pi/2} + 2 \cdot I_{1,2}  \r \\
&= \frac13 - \frac15 \l 1 + \frac23 \l \left [-\cos x \cos 2 x \right]_0^{\pi/2} + 1\cdot I_{0,1} \r  \r \\
&= \frac13 - \frac15 -\frac2{15} - \frac2{15} \\
&= \frac{5}{15} - \frac{3}{15} - \frac{4}{15} \\
&= -\frac2{15}
\end{align*}
\end{questionparts}