How do I solve using substitution?
$\displaystyle \int\sin4xtanxdx$
Thanks for your help.
Consider the following two identities:
$\displaystyle cos(a - b) = cos(a)cos(b) + sin(a)sin(b)$
$\displaystyle cos(a + b) = cos(a)cos(b) - sin(a)sin(b)$
Subtract the bottom from the top:
$\displaystyle cos(a - b) - cos(a + b) = 2sin(a)sin(b)$
Thus
$\displaystyle sin(a)sin(b) = \frac{1}{2}(cos(a - b) - cos(a + b))$
Thus
$\displaystyle sin(4x)sin(x) = \frac{1}{2}(cos(4x - x) - cos(4x + x))$
$\displaystyle sin(4x)sin(x) = \frac{1}{2}(cos(3x) - cos(5x))$
-Dan