Area of Region Under Graph x=-pi/6 to x=pi/4: Find the Solution

[tomwilliam]

Homework Statement

Find the area of the region between x=-pi/6 and x=pi/4 above the x-axis and below the graph of:
y=cos(2x)(8-cos(2x))

Homework Equations

The Attempt at a Solution

I know I need to reformulate the expression somehow so that it is easier to integrate, and I'm guessing it's a double angle formula like:
cos (2x) = 2 cos^2 x -1
but I can't seem to bring it all together to find something I know how to integrate.
Any hints? This is an assignment question, so I'd like to know how to do it properly. I've also changed a few of the values (but not the argument of the trig functions).

 

[Whitishcube]

good. so far youre off to a good start. use that trig identity and play around with it once you substitute that in your integral.

 

[ehild]

Homework Statement

Find the area of the region between x=-pi/6 and x=pi/4 above the x-axis and below the graph of:
y=cos(2x)(8-cos(2x))

Homework Equations

The Attempt at a Solution

I know I need to reformulate the expression somehow so that it is easier to integrate, and I'm guessing it's a double angle formula like:
cos (2x) = 2 cos^2 x -1

You guess well, but use the formula in the opposite way, to make cos^2 (2x) disappear. You can integrate cos2x and cos4x.

ehild

 

[tomwilliam]

I'm a little confused by that...if I use the formula the opposite way to get cos^2 to disappear, I get back to what I started with.
At the moment I've changed the cos(2x) and expanded the brackets to get:

16 cos^2 (x) - 7 - 4 cos^4 (x)

I'm not sure if I'm any closer or not.

 

[l'Hôpital]

You are thinking about it the wrong way. You have

f(x) = 8 cos(2x) - cos^2 (2x)

You could easily do u sub, so 8 cos(u) - cos^2(u). Integrating cos is easy, but what about cos^2?

 

[tomwilliam]

Ah yes, I see now.
Thanks all.