Integrals Involving Trig Functions — Question 10

Question 10 – Solution

First expand the square: $$\begin{align*} E &=\int_0^{2\pi}\bigl(\sin(3t)+\sin(5t)\bigr)^2\,dt\\ &=\int_0^{2\pi}\left[\sin^2(3t) +2\sin(3t)\sin(5t)+\sin^2(5t)\right]dt. \tag{1} \end{align*}$$ For the cross term, use $$2\sin(A)\sin(B)=\cos(A-B)-\cos(A+B).$$ Thus, $$\begin{align*} 2\int_0^{2\pi}\sin(3t)\sin(5t)\,dt &=\int_0^{2\pi}\bigl(\cos(2t)-\cos(8t)\bigr)\,dt\\ &=\left[\frac12\sin(2t)-\frac18\sin(8t)\right]_0^{2\pi}=0. \end{align*}$$ For each squared term, use $\sin^2(kt)=\frac12(1-\cos(2kt))$: $$\begin{align*} \int_0^{2\pi}\sin^2(kt)\,dt &=\frac12\int_0^{2\pi}\bigl(1-\cos(2kt)\bigr)\,dt\\ &=\frac12\left[t-\frac{\sin(2kt)}{2k}\right]_0^{2\pi}=\pi \end{align*}$$ for any positive integer $k$. In particular, the $k=3$ and $k=5$ terms each contribute $\pi$. Substituting into equation (1), $$E=\pi+0+\pi=2\pi.$$ Therefore, $$\boxed{E=2\pi}.$$