Evaluate the integral.
-$\int \sin 8 t \sin 5 \mathrm { tdt }$
A) $\frac { 1 } { 6 } \sin 8 t - \frac { 1 } { 26 } \sin 5 t + C$
B) $\frac { 1 } { 6 } \sin 3 t - \frac { 1 } { 26 } \cos 13 t + C$
C) $\frac { 1 } { 6 } \sin 3 t + \frac { 1 } { 26 } \sin 13 t + C$
D) $\frac { 1 } { 6 } \sin 3 t - \frac { 1 } { 26 } \sin 13 t + C$

Question

Quizplus · Accepted Answer

Using the identity $\sin a \sin b = \frac{1}{2} (\cos(a-b) - \cos(a+b))$, we have 
\begin{align*}
\int \sin 8 t \sin 5 t \ dt &= \frac{1}{2} \int (\cos(8t-5t) - \cos(8t+5t)) \ dt \
&= \frac{1}{2} \int (\cos 3t - \cos 13t) \ dt \
&= \frac{1}{6} \sin 3t - \frac{1}{26} \sin 13t + C.
\end{align*} 
Thus, the correct choice is (D).

Evaluate the Integral ∫sin⁡8tsin⁡5tdt\int \sin 8 t \sin 5 \mathrm { tdt }∫sin8tsin5tdt

Evaluate the Integral $\int \sin 8 t \sin 5 \mathrm { tdt }$