Evaluate the integral...?

Question

evaluate the integral from (pi/3 to pi/2)
sinx√(9-36cos²x)dx ; use u = 6cosx ?

Give answer in decimal form. Thanks so much for the help :)

Answer 1

du=- 6sinxdx
sinxdx (9-36cos^2x)^1/2 = (- du/6) (9 - u^2)^1/2

if dv=du/6 and t= -(9-u^2)^1/2 then

int( (- du/6) (9 - u^2)^1/2)pi/3 to pi/2=[ u/6 *(-(9-u^2)^1/2) ] pi/3 to pi/2 - int( u/6 * D(t))pi/3 to pi/2

Evaluate the terms in the right...
Actually I used the following formula:
int(vdt)a to b = vt]a tob - int(tdv)a to b
=v(a)t(a) - v(b)t(b) - int(tdv)a to b

you may need to use this formula again....

Answer 2

Substitute u = 6cosx
du = -6 sinx dx
and:
√(9-36cos²x) = √(3^2 - u^2)

Then sinx√(9-36cos²x)dx = -(1/6)√(3^2 - u^2) du

(Integration by parts as below. There may be some neat substitution here, along the lines of: tan ax = √1 - sec^2 ax )

Anyway:
Integ_pi/3_pi/2 [ sinx√(9-36cos²x)dx ]
= -(1/6) Integ_pi/3_pi/2 [√(3^2 - u^2) du ]

Integrate by parts: Integ SdT = [ST] - Integ TdS
Setting: S=√(3^2 - u^2) , dT=du
Then dS = -2udu/2√(3^2 - u^2) , T = u (+ C)
Integ SdT = [ST] - Integ TdS = (√(3^2 - u^2))u - Integ [udu/2√(3^2 - u^2)]

Integ [-udu/√(3^2 - u^2)] is one of the standard forms, like cos ax -> cot ax.