First we address the special case where $m = n$. In this case $m + n = m + m = 2 m$ is even. We have: {{begin-eqn}} {{eqn | l = \int \sin m x \cos m x \rd x       | r = \frac {\sin^2 m x} {2 m} + C       | c = Primitive of $\sin m x \cos m x$ }} {{eqn | ll= \leadsto       | l = \int_0^\pi \sin m x \cos m x \rd x       | r = \left[{\frac {\sin^2 m x} {2 m} }\right]_0^\pi       | c =  }} {{eqn | r = \frac {\sin^2 m \pi} {2 m} - \frac {\sin^2 0} {2 m}       | c =  }} {{eqn | r = 0       | c = Sine of Multiple of Pi }} {{end-eqn}} So in this case, $\displaystyle \int_0^\pi \sin m x \cos n x \rd x = 0$ for $m + m$ even. Let $m \ne n$. {{begin-eqn}} {{eqn | l = \int \sin m x \cos n x \rd x       | r = \frac {-\cos \left({m - n}\right) x} {2 \left({m - n}\right)} - \frac {\cos \left({m + n}\right) x} {2 \left({m + n}\right)} + C       | c = Primitive of $\sin m x \cos n x$ }} {{eqn | ll= \leadsto       | l = \int_0^\pi \sin m x \cos n x \rd x       | r = \left[{\frac {-\cos \left({m - n}\right) x} {2 \left({m - n}\right)} - \frac {\cos \left({m + n}\right) x} {2 \left({m + n}\right)} }\right]_0^\pi       | c =  }} {{eqn | r = \left({\frac {-\cos \left({\left({m - n}\right) \pi}\right)} {2 \left({m - n}\right)} - \frac {\cos \left({\left({m + n}\right) \pi}\right)} {2 \left({m + n}\right)} }\right)       | c =  }} {{eqn | o =        | ro= -       | r = \left({\frac {-\cos 0} {2 \left({m - n}\right)} - \frac {\cos 0} {2 \left({m + n}\right)} }\right)       | c =  }} {{eqn | r = \frac {-\cos \left({\left({m - n}\right) \pi}\right)} {2 \left({m - n}\right)} - \frac {\cos \left({\left({m + n}\right) \pi}\right)} {2 \left({m + n}\right)} + \frac 1 {2 \left({m - n}\right)} + \frac 1 {2 \left({m + n}\right)}       | c = Cosine of Zero is One and simplifying }} {{eqn | r = \frac 1 {2 \left({m - n}\right)} + \frac 1 {2 \left({m + n}\right)} - \frac {\left({-1}\right)^{m - n} } {2 \left({m - n}\right)} - \frac {\left({-1}\right)^{m + n} } {2 \left({m + n}\right)}       | c = Cosine of Multiple of Pi and rearranging }} {{eqn | r = \frac 1 {2 \left({m - n}\right)} + \frac 1 {2 \left({m + n}\right)} - \frac {\left({-1}\right)^{m + n} } {2 \left({m - n}\right)} - \frac {\left({-1}\right)^{m + n} } {2 \left({m + n}\right)}       | c = $m + n$ and $m - n$ have the same parity }} {{end-eqn}} {{qed|lemma}} When $m + n$ is an even integer, we have: {{begin-eqn}} {{eqn | o =        | r = \frac 1 {2 \left({m - n}\right)} + \frac 1 {2 \left({m + n}\right)} - \frac {\left({-1}\right)^{m + n} } {2 \left({m - n}\right)} - \frac {\left({-1}\right)^{m + n} } {2 \left({m + n}\right)}       | c =  }} {{eqn | r = \frac 1 {2 \left({m - n}\right)} + \frac 1 {2 \left({m + n}\right)} - \frac 1 {2 \left({m - n}\right)} - \frac 1 {2 \left({m + n}\right)}       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} This shows that in all cases $\displaystyle \int_0^\pi \sin m x \cos n x \rd x = 0$ for $m + m$ even. When $m + n$ is an odd integer, we have: {{begin-eqn}} {{eqn | o =        | r = \frac 1 {2 \left({m - n}\right)} + \frac 1 {2 \left({m + n}\right)} - \frac {\left({-1}\right)^{m + n} } {2 \left({m - n}\right)} - \frac {\left({-1}\right)^{m + n} } {2 \left({m + n}\right)}       | c =  }} {{eqn | r = \frac 1 {2 \left({m - n}\right)} + \frac 1 {2 \left({m + n}\right)} - \frac {-1} {2 \left({m - n}\right)} - \frac {-1} {2 \left({m + n}\right)}       | c =  }} {{eqn | r = \frac 1 {m - n} + \frac 1 {m + n}       | c = simplification }} {{eqn | r = \frac {\left({m + n}\right) + \left({m - n}\right)} {\left({m + n}\right) \left({m - n}\right)}       | c =  }} {{eqn | r = \frac {2 m} {\left({m + n}\right) \left({m - n}\right)}       | c =  }} {{eqn | r = \frac {2 m} {m^2 - n^2}       | c = Difference of Two Squares }} {{end-eqn}} {{qed}}