Answer
$\int_0^2\sin (2\pi x) \cos (5\pi x) dx = 0$
Work Step by Step
It follows from the graph that $\int_0^2\sin (2\pi x) \cos (5\pi x) dx = 0$, since the bulge above the x-axis appears to have a corresponding depression below the x-axis.
$I=\int_0^2\sin (2\pi x) \cos 5\pi x dx$
Using:
$\sin u\cos v=\frac{1}{2}\left [ \sin(u+v)+\sin(u-v) \right ] \quad$
$=\int_0^2\frac{1}{2}\left [ \sin(2\pi x+5\pi x)+\sin(2\pi x-5\pi x) \right ]dx \quad$
$=\int_0^2\frac{1}{2}\left [ \sin(7\pi x)+\sin(-3\pi x) \right ]dx \quad$
$=\frac{1}{2} \left [\int_0^{2} \sin(7\pi x)dx +\int_0^{2} \sin(-3\pi x) dx\right ] \quad $
$\begin{Bmatrix}
w=7\pi x & \\
dw=7\pi dx & \frac{dw}{7\pi }=dx
\end{Bmatrix}$
$\begin{Bmatrix}
m=-3\pi x & \\
dw=-3\pi dx & \frac{dw}{-3\pi }=dx
\end{Bmatrix}$
$=\frac{1}{2} \left [\int_0^{14\pi} \sin(w) \frac{dw}{7\pi } +\int_0^{-6\pi} \sin(m) \frac{dw}{-3\pi }\right ] \quad $
$=\frac{1}{2} \left [\int_0^{14\pi} \sin(w) \frac{dw}{7\pi } -\int_{-6\pi}^{0} \sin(m) (-\frac{dw}{3\pi })\right ] \quad $
$=\frac{1}{14\pi} \left [(-\cos w) \right ]_0^{14\pi} \quad +\frac{1}{6\pi} \left [(-\cos w) \right ]_{-6\pi}^{0} \quad $
$=\frac{1}{14\pi} (-1+1) +\frac{1}{6\pi}(-1+1)$
$=0$
