Hi,

Can someone please show me how to solve this, step by step?

Integrate the following:sinxcosxcos2x

Apparently the answer is1/8(sin2x)^2 + C

Does sinxcosx go outside of the integrand as a factor?

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- Apr 29th 2010, 03:24 PMCSG18Integration 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? - Apr 29th 2010, 04:11 PMskeeter
- Apr 30th 2010, 10:52 AMCSG18
- Apr 30th 2010, 11:01 AMharish21
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 - May 1st 2010, 03:05 AMHallsofIvy
- May 1st 2010, 07:19 AMCSG18
- May 1st 2010, 07:32 AMe^(i*pi)
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