Asked by Bae
                Integrate the following:
a. Integral from 0 to pi (sin^2)(3x)dx
b. Integral of (x^2)/((x^2 - 4)^3/2)
            
        a. Integral from 0 to pi (sin^2)(3x)dx
b. Integral of (x^2)/((x^2 - 4)^3/2)
Answers
                    Answered by
            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
    
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
                    Answered by
            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^2-4) + x)) - x/√(x^2-4) + 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%5E2-4%29+%2B+x%29%29+-+x%2F√%28x%5E2-4%29
    
Gave up and used Wolfram with results of
log(2(√(x^2-4) + x)) - x/√(x^2-4) + 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%5E2-4%29+%2B+x%29%29+-+x%2F√%28x%5E2-4%29
                    Answered by
            bobpursley
            
    I tried change of variables (trig functions) on number 2, and got messy also, gave up.
    
                    Answered by
            Steve
            
    ∫(x^2)/((x^2 - 4)^3/2 dx
x = 2coshθ
dx = 2sinhθ dθ
x^2-4 = 4sinh^2θ
∫4cosh^2θ/8sinh^3θ 2sinhθ dθ
Keeping in mind that
arccoshθ = log(z+√(z^2-1))
I think wolfram's answer is less mysterious.
Just as trig substitutions are your friend, so are the hyperbolic functions.
    
x = 2coshθ
dx = 2sinhθ dθ
x^2-4 = 4sinh^2θ
∫4cosh^2θ/8sinh^3θ 2sinhθ dθ
Keeping in mind that
arccoshθ = log(z+√(z^2-1))
I think wolfram's answer is less mysterious.
Just as trig substitutions are your friend, so are the hyperbolic functions.
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