Posted by Bae on Friday, May 2, 2014 at 8:25pm.
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, Friday, May 2, 2014 at 9:44pm
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, Friday, May 2, 2014 at 9:56pm
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
Calc 2 - bobpursley, Friday, May 2, 2014 at 10:00pm
I tried change of variables (trig functions) on number 2, and got messy also, gave up.
Calc 2 - Steve, Saturday, May 3, 2014 at 12:14am
∫(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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