what's the approach used in this integral? in : $\displaystyle \int (\sin 3x)(\cos 5x)\, dx $ ( which recurred as hyperbolic sin and cosine as well ) 


*

*ive noted the resemblance to the sum of $ \sin 8x$ but didnt know how to use it 

*ive noted that $ \cos 5x $ can be expanded to $ \cos (2x+3x ) $ but again failed to simplify the resulted form


*

*Someone already suggested using the identity $$\sin A \cos B = \frac{ \sin(A-B) + \sin(A+B)} 2.$$ That's one way.

*Another way is this:
\begin{align}
& \int \sin(3x)\cos(5x)\,dx = \int \frac{e^{3ix} - e^{-3ix}}{2i} \cdot \frac{e^{5ix}+ e^{-5ix}} 2 \,dx \\[10pt]
= {} & \frac 1 {4i} \int (e^{8ix} - e^{2ix} - e^{-2ix} + e^{-8ix}) \,dx
\end{align}
then integrate term by term and then use $e^{i\theta} = \cos\theta + i\sin\theta$ to get a function of sines and cosines.

*And here's another way:
\begin{align}
& \int \sin(3x)\Big(\cos(5x)\,dx\Big) = \overbrace{\int u\,dv = uv - \int v\,du}^\text{integration by parts} \\[10pt]
= {} & \sin(3x) \frac{\sin(5x)} 5 - \int \frac {\sin(5x)} 5 \cdot 3\cos(3x)\,dx \\[10pt]
= {} & \frac 1 5 \sin(3x)\sin(5x) - \frac 3 5 \int \cos(3x) \Big(\sin(5x)\,dx\Big) \\[10pt]
= {} & \frac 1 5 \sin(3x)\sin(5x) - \frac 3 5 \left( \int s\,dt \right) \\[10pt]
= {} & \frac 1 5 \sin(3x)\sin(5x) - \frac 3 5 \left( st - \int t \,ds \right) \\[10pt]
= {} & \frac 1 5 \sin(3x)\sin(5x) - \frac 3 5 \left( \frac{-1} 5 \cos(3x)\cos(5x) - \int \frac 3 5 \sin(3x)\cos(5x)\,dx \right) \\[10pt]
= {} & \frac 1 5 \sin(3x)\sin(5x) + \frac 3 {25} \cos(3x)\cos(5x) + \frac 9 {25} \int \sin(3x)\cos(5x)\, dx \\[10pt]
& \text{and so we have:} \\
I = {} &  \frac 1 5 \sin(3x)\sin(5x) + \frac 3 {25} \cos(3x)\cos(5x) + \frac 9 {25} I. \\[10pt]
& \text{Adding } \frac 9 {25} I \text{ to both sides, we get:} \\
\frac{34}{25} I = {} & \frac 1 5 \sin(3x)\sin(5x) + \frac 3 {25} \cos(3x)\cos(5x) + \text{constant} \\[10pt]
& \text{and then multiply both sides by } 25/34.
\end{align}
