Tuesday

January 27, 2015

January 27, 2015

Posted by **Bae** on Friday, May 2, 2014 at 8:25pm.

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 9:44pma) (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**, Friday, May 2, 2014 at 9:56pmtried 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

- Calc 2 -
**bobpursley**, Friday, May 2, 2014 at 10:00pmI 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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