1. ## Integration with trigonometry

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?

2. Originally Posted by CSG18
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 $\sin{2X} = 2\sin{X}\cos{X}$,

$\cos{2X} = \cos^2{X} - \sin^2{X}$

and $\sin^2{x} + \cos^2{x} = 1$.

That means

$\sin{x}\cos{x}\cos{2x} = \frac{1}{2}\sin{2x}\cos{2x}$

$= \frac{1}{4}\sin{4x}$.

Therefore:

$\int{\sin{x}\cos{x}\cos{2x}\,dx} = \int{\frac{1}{4}\sin{4x}\,dx}$

$= -\frac{1}{16}\cos{4x} + c$

$= -\frac{1}{16}(\cos^2{2x} - \sin^2{2x}) + c$

$= -\frac{1}{16}(1 - \sin^2{2x} - \sin^2{2x}) + c$

$= -\frac{1}{16}(1 - 2\sin^2{2x}) + c$

$= -\frac{1}{16} + \frac{1}{8}\sin^2{2x} + c$

$= \frac{1}{8}\sin^2{2x} + C$, where $C = c - \frac{1}{16}$.