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
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
tested it and got the original back after differentiating it
I tried change of variables (trig functions) on number 2, and got messy also, gave up.
∫(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.
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