Question 13.6: Find (a) ∫ sin 2t cos t dt (b) ∫ sin mt sin nt dt, where m a......

(a) Using the identities in Table 3.1 we find

$2 \sin A \cos B=\sin (A+B)+\sin (A-B)$

hence  $\sin 2 t \cos t$  can be written  $\frac{1}{2}(\sin 3 t+\sin t)$.  Therefore,

$\begin{aligned}\int \sin 2 t \cos t \mathrm{~d} t & =\int \frac{1}{2}(\sin 3 t+\sin t) \mathrm{d} t \\& =\frac{1}{2}\left(\frac{-\cos 3 t}{3}-\cos t\right)+c \\& =-\frac{1}{6} \cos 3 t-\frac{1}{2} \cos t+c\end{aligned}$

(b) Using the identity  $2 \sin A \sin B=\cos (A-B)-\cos (A+B)$,  we find

$\sin m t \sin n t=\frac{1}{2}\{\cos (m-n) t-\cos (m+n) t\}$

Therefore,