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
Integrate the following:
a. Integral from 0 to pi (sin^2)(3x)dx
b. Integral of (x^2)/((x^2 - 4)^3/2)
4 answers
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
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.
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.
0 answers