Evaluate the Definitel Integral...?

Question

Evaluate the definitely integral by expressing it in terms of u and evaluating the resulting integral using formulas from geometry.

The integral is from (pi/3 to pi/2) and it is of

sin(theta)root(9-36cossquaredtheta) dtheta

use u =6costheta

Thanks so much for help everybody! Oh and decimal answers are acceptable. thanks. :)

it is sin(x) times the square root of the next term...
I used x for theta.

hold your mouse over the equation to see the whole thing.

Answer 1

is it sin@ over the root or it's multiplied?
Because if it is over it then you could solve it otherwise you couldn't I Guess!
Integral (Pi/3 to Pi/2) Sin@ / Root(9-36Cos^2@) d@
This is the way I solved it:
u=6Cos@ a=3
du= -6Sin@ d@
Multiply both sides by (-1/6Sin@) Then you get:
-du/6Sin@=d@
Then go plug these in to you original equation:
You Get: Intg (Pi/3 to Pi/2) (Sin@/Root(a^2-u^2))*(-du/6Sin@)
Then Sin@ cancel out and you can also pull the constant out which is (-1/6) out of your Integral Then you'll get:
(-1/6)*Intg (Pi/3 to Pi/2) 1/Root(a^2-u^2)*du
Then you know that the formula of the Derivative of ArcSin@ is: du/Root(a^2-u^2)
and the Antiderevative of ArcSin@ is: ArcSin(u/a) + C
So: (-1/6)*ArcSin((6Cos@)/3)

Then you Plug in your intervals;
(-1/6)*ArcSin(2Cos(Pi/2)) - (-1/6)*ArcSin(2Cos(Pi/3))

If you plug in this in your Calculator you'll get the right answer I hope :)
Answer is around: .2618