Question$\int \sin 5 x \cdot \cos 3 x d x$

Studdy Solution
Separate the integral into two parts and integrate each term individually:
$= \frac{1}{2} \left( \int \sin(8x) \, dx + \int \sin(2x) \, dx \right)$
Integrate each sine function:
1. For $\int \sin(8x) \, dx$, use the substitution $u = 8x$, $du = 8 \, dx$, hence $dx = \frac{1}{8} du$:
$\int \sin(8x) \, dx = -\frac{1}{8} \cos(8x) + C_1$
2. For $\int \sin(2x) \, dx$, use the substitution $v = 2x$, $dv = 2 \, dx$, hence $dx = \frac{1}{2} dv$:
$\int \sin(2x) \, dx = -\frac{1}{2} \cos(2x) + C_2$
Combine the results:
$= \frac{1}{2} \left( -\frac{1}{8} \cos(8x) - \frac{1}{2} \cos(2x) \right) + C$
Simplify:
$= -\frac{1}{16} \cos(8x) - \frac{1}{4} \cos(2x) + C$
The solution to the integral is:
$\boxed{-\frac{1}{16} \cos(8x) - \frac{1}{4} \cos(2x) + C}$