Hi,
Can someone please show me how to solve this, step by step?
Integrate the following: sinxcosxcos2x
Apparently the answer is 1/8(sin2x)^2 + C
Does sinxcosx go outside of the integrand as a factor?
These sorts of problems are easiest if you can use trigonometric identities to simplify the integrand.
Remember that $\displaystyle \sin{2X} = 2\sin{X}\cos{X}$,
$\displaystyle \cos{2X} = \cos^2{X} - \sin^2{X}$
and $\displaystyle \sin^2{x} + \cos^2{x} = 1$.
That means
$\displaystyle \sin{x}\cos{x}\cos{2x} = \frac{1}{2}\sin{2x}\cos{2x}$
$\displaystyle = \frac{1}{4}\sin{4x}$.
Therefore:
$\displaystyle \int{\sin{x}\cos{x}\cos{2x}\,dx} = \int{\frac{1}{4}\sin{4x}\,dx}$
$\displaystyle = -\frac{1}{16}\cos{4x} + c$
$\displaystyle = -\frac{1}{16}(\cos^2{2x} - \sin^2{2x}) + c$
$\displaystyle = -\frac{1}{16}(1 - \sin^2{2x} - \sin^2{2x}) + c$
$\displaystyle = -\frac{1}{16}(1 - 2\sin^2{2x}) + c$
$\displaystyle = -\frac{1}{16} + \frac{1}{8}\sin^2{2x} + c$
$\displaystyle = \frac{1}{8}\sin^2{2x} + C$, where $\displaystyle C = c - \frac{1}{16}$.