solve: (1) integral of (sin7xcos3x)dx (2) integral of (cos3xcos4x)dx?

Answer 1

(sin7xcos3x)
multiply and divide by 2
2(sin7xcos3x)/2
use formula
2sinAcosB
=sin(A+B)+sin(A-B)
it becomes
(sin10x+sin4x)/2
now integrate u get
[(-cos10x)/10+(-cos4x)/4]/2

for (2)part
use formula
2cosAcosB
= cos(A+B)+cos(A-B)