We have the following integral:
[tex]\int sin(4x)cos(24x)dx[/tex]
Changing variables:
[tex]u=4x \rightarrow dx=\frac{1}{4}dx[/tex]
Therefore:
[tex]\frac{1}{4}\int sin(u)cos(6u)du[/tex]
From trigonometrical identities:
[tex]sin(\alpha)cos(\beta)= \frac{sin(\beta+\alpha)-sin(\beta-\alpha)}{2}[/tex]
Thus:
[tex]\frac{1}{4}\int \frac{sin(7u)-sin(5u)}{2}du= \frac{1}{8} \int sin(7u)du-\frac{1}{8} \int sin(5u)du[/tex]
∴ [tex]\frac{-cos(7u)}{56}+ \frac{cos(5u)}{40}+C[/tex]
[tex]u \rightarrow 4x[/tex]
Finally:
[tex]\boxed{ I= \frac{cos(20x)}{40}-\frac{cos(28x)}{56}+C}[/tex]
