Posted by Bae on .
Integrate the following:
a. Integral from 0 to pi (sin^2)(3x)dx
b. Integral of (x^2)/((x^2  4)^3/2)

Calc 2 
Reiny,
a) (sin^2)(3x)
using cos 2A = 1  2sin^2 A
cos 6x = 1  2sin^2 (3x)
sin^2 3x = 1/2  (1/2) cos 6x
so integral sin^2 3x = (1/2)x  (1/12)sin(6x)
Take it from there, I will let you do the substitution and evaluation 
Calc 2 
Reiny,
tried integration by parts on the 2nd, but it got messy, probably made an error
Gave up and used Wolfram with results of
log(2(√(x^24) + x))  x/√(x^24) + C
http://integrals.wolfram.com/index.jsp?expr=%28x%5E2%29%2F%28%28x%5E2++4%29%5E%283%2F2%29%29&random=false
tested it and got the original back after differentiating it
http://www.wolframalpha.com/input/?i=derivative+log%282%28√%28x%5E24%29+%2B+x%29%29++x%2F√%28x%5E24%29 
Calc 2 
bobpursley,
I tried change of variables (trig functions) on number 2, and got messy also, gave up.

Calc 2 
Steve,
∫(x^2)/((x^2  4)^3/2 dx
x = 2coshθ
dx = 2sinhθ dθ
x^24 = 4sinh^2θ
∫4cosh^2θ/8sinh^3θ 2sinhθ dθ
Keeping in mind that
arccoshθ = log(z+√(z^21))
I think wolfram's answer is less mysterious.
Just as trig substitutions are your friend, so are the hyperbolic functions.