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?
Because $\displaystyle sin(2x) = 2 sinx cosx$
If you dont know this, you need to look back at the topic double angle formula. Here it is:
Double Angle
The fraction is there because skeeter multiplied the term $\displaystyle \sin{x}\cos{x}\cos(2x)$ by $\displaystyle \frac{2}{2}$ , which is equal to 1
No because $\displaystyle \cos(x)$ is not a constant. 4 can go out of the integrand but this is not the optimal way to solve it
If you use skeeter's hint of $\displaystyle u = \sin(2x)$ then $\displaystyle du = 2\cos(2x)\,dx$
Therefore $\displaystyle \int \cos(2x) \sin(2x)\,dx = \frac{1}{2}\int u\,du$
Note that I have taken a factor of 1/2 out of the integral - this is because 1/2 is a constant