Integrate the following:
a. Integral from 0 to pi (sin^2)(3x)dx
b. Integral of (x^2)/((x^2 - 4)^3/2)
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
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
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.
To integrate the given expressions, we will need to apply appropriate integration techniques. Let's solve each integral step by step:
a. Integral from 0 to pi of (sin^2)(3x)dx:
To evaluate this integral, we can use the trigonometric identity sin^2(x) = (1 - cos(2x))/2.
So, rewriting the integral with this identity, we have:
∫(0 to pi) (1 - cos(6x))/2 dx
Now, let's split the integral into two parts:
∫(0 to pi) (1/2) dx - ∫(0 to pi) (cos(6x)/2) dx
The first part is straightforward:
(1/2) ∫(0 to pi) dx = (1/2) [x] from 0 to pi = (1/2) * (pi - 0) = pi/2
For the second part, we deal with the integral of cos(6x):
∫(0 to pi) (cos(6x)/2) dx = (1/2) ∫(0 to pi) cos(6x) dx
To integrate cos(mx), we use the formula:
∫cos(mx) dx = (1/m) sin(mx)
Applying the formula, we get:
(1/2) ∫(0 to pi) cos(6x) dx = (1/2) * [(1/6) sin(6x)] from 0 to pi
= (1/2) * [(1/6) sin(6pi) - (1/6) sin(0)]
= (1/2) * [(1/6) * 0 - (1/6) * 0]
= 0
So, the value of the integral is pi/2.
b. Integral of (x^2)/((x^2 - 4)^3/2):
To solve this integral, we can use a substitution. Let's substitute u = x^2 - 4, which gives us du = 2x dx.
Now, rewrite the integral using u and du:
∫(x^2)/((x^2 - 4)^3/2) dx = (1/2) ∫(1/u^3/2) du
Let's simplify the expression:
(1/2) ∫u^(-3/2) du = (1/2) [2u^(-1/2)] + C
= u^(-1/2) + C
Finally, substituting the original variable back in:
= (x^2 - 4)^(-1/2) + C
So, the integral of (x^2)/((x^2 - 4)^3/2) is (x^2 - 4)^(-1/2) + C.