This is my assignment.
I need to find the dx of this integral, and show the integration process. I have no idea how to start, let alone do it.
Integral of (the sybol has a zero on the bottom, and a 3 at the top) the square root of (9-x^2) dx
The teacher gave us the following extra info:
x = 3 sin u
(It becomes) Integral of the square root of (9 - 9sin^2 u)
The answer is 9pi/4
I need to find dx and show the integration process
okay so, you get the integral of the square root of 9 (1 - sin^2 u) dx
or the integral of 3 times the square root of (1 - sin^2 u) dx
or the integral of 3 times the square root of (cos^2 u) dx
or just the integral of 3 cos u dx.
Okay, but, we need to find du if we want to figure out dx in terms of du - to do that we just take the derivative of 3 sin u, so dx/du = 3 cos u, or dx = 3 cos u du.
So, we have the integral of (3 cos u)^2 du
But we need to also convert the bounds to the “u” values we’re using, so for the lower bound, 0 = 3 sin u, u = 0
For the upper bound, 3 = 3 sin u, u = pi/2.
So integral (0, pi/2) of (3 cos u)^2 du = 9 times integral (0, pi/2) of cos^2 u
and errrr, I forget exactly how to solve that, which maybe means I did something wrong but I don’t think so. It’s been awhile since I’ve done calculus though. You can probably get away with only that much work though.
following Mazrim Taim’s calculations (which seem correct):
we have 9 times the integral cos^2(u) du (between 0 and pi/2)
we know that cos^2 u = (1 + cos(2u))/2
so, replacing that on our integral we get:
9 * (integral (0, pi/2) of (1 + cos(2u))/2 du) <=>
9 * (integral (0, pi/2) of (1/2 + cos(2u)/2) du)
we integrate the expression and get:
9 * (u/2 + (sen(2u))/4) between 0 and pi/2
now we use the first fundamental teorem and get:
(9 * ((pi/2)/2 + (sen(2*(pi/2))))) - (9 * (0/2 +sen(2*0))) <=>
note: sin 0 = 0;
(9 * (pi/4 + sen(pi)) - (9 * (0 + 0)) <=>
note: sin pi = 0;
9*(pi/4) <---- the answer
ah yeah, pulling the old (1 + cos (2u))/2 trick. clever.