could someone check ive done this right.... many thnaks

evaluate

$\displaystyle

\int {sin (4x)}{cos (x)} dx

$

i have....

$\displaystyle \int\frac{1}{2} {sin (a+b)}{sin (a-b)} dx

$

$\displaystyle \int\frac{1}{2} {sin (5x)}{sin (3x)} dx

$

$\displaystyle \int\frac{1}{2}\frac {(-cos (5x))}{5}+cos (3x)+c dx

$