# Thread: Integration with trigonometry

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?

only constants can be factored out of an integral ... you know that.
$\sin{x}\cos{x}\cos(2x) =$

$\frac{1}{2} \cdot 2\sin{x}\cos{x}\cos(2x) =$

$\frac{1}{2} \cdot \sin(2x)\cos(2x)$

substitution ... let $u = \sin(2x)$

3. Originally Posted by skeeter
$\sin{x}\cos{x}\cos(2x) =$

$\frac{1}{2} \cdot 2\sin{x}\cos{x}\cos(2x) =$

$\frac{1}{2} \cdot \sin(2x)\cos(2x)$

substitution ... let $u = \sin(2x)$
Sorry I do not understand what you did...

How did it turn into $sin(2x)\cos(2x)$ ?

I understand where the fraction came from though.

4. Originally Posted by CSG18
Sorry I do not understand what you did...

How did it turn into $sin(2x)\cos(2x)$ ?

I understand where the fraction came from though.
Because $sin(2x) = 2 sinx cosx$

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 $\sin{x}\cos{x}\cos(2x)$ by $\frac{2}{2}$ , which is equal to 1

5. 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?
A function of the variabel NEVER "goes outside the integrand"! Only constants (with respect to the variable of integration) can be moved like that.

Skeeter used the trig identity that sin(2x)= 2sin(x)cos(x) so that sin(x)cos(x)= (1/2)sin(2x).

6. Originally Posted by HallsofIvy
A function of the variabel NEVER "goes outside the integrand"! Only constants (with respect to the variable of integration) can be moved like that.

So why is it when you integrate 4cosx sin^3x dx, 4 cosx goes outside as a factor?

7. Originally Posted by CSG18
So why is it when you integrate 4cosx sin^3x dx, 4 cosx goes outside as a factor?
No because $\cos(x)$ 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 $u = \sin(2x)$ then $du = 2\cos(2x)\,dx$

Therefore $\int \cos(2x) \sin(2x)\,dx = \frac{1}{2}\int u\,du$

Note that I have taken a factor of 1/2 out of the integral - this is because 1/2 is a constant