Integrals Involving Trig Functions — Question 7

Question 7 – Solution

Use $$\sin(A)\cos(B)=\frac12\bigl[\sin(A+B)+\sin(A-B)\bigr].$$ Setting $A=mx$ and $B=nx$ gives $$\sin(mx)\cos(nx) =\frac12\bigl[\sin((m+n)x)+\sin((m-n)x)\bigr]. \tag{1}$$ For every nonzero integer $k$, $$\begin{align*} \int_0^{2\pi}\sin(kx)\,dx &=\left[-\frac{\cos(kx)}k\right]_0^{2\pi}\\ &=\frac{-\cos(2\pi k)+\cos(0)}k=0, \end{align*}$$ because $\cos(2\pi k)=1$. Since $m+n\geq2$, the first sine term in equation (1) always integrates to zero.

If $m\ne n$, then $m-n$ is a nonzero integer, so the second term also integrates to zero. If $m=n$, then $$\sin((m-n)x)=\sin(0)=0,$$ so it again contributes zero. Thus, in every case, $$\int_0^{2\pi}\sin(mx)\cos(nx)\,dx=0.$$ Therefore, there are no integers $m,n\geq1$ for which the integral is nonzero: $$\boxed{\text{There are no such integer pairs }(m,n).}$$