I need help.

Compute the definite integral:
 integral_(-3)^3 sqrt(9-x^2) dx
For the integrand sqrt(9-x^2), substitute x = 3 sin(u) and  dx = 3 cos(u)  du.
Then sqrt(9-x^2)  =  sqrt(9-9 sin^2(u))  =  3 sqrt(cos^2(u)).
This substitution is invertible over -pi/2<u<pi/2 with inverse u = sin^(-1)(x/3).
This gives a new lower bound u = sin^(-1)(-3/3) = -pi/2 and upper bound u = sin^(-1)(3/3) = pi/2:
  =  3 integral_(-pi/2)^(pi/2) 3 cos(u) sqrt(cos^2(u)) du
Factor out constants:
  =  9 integral_(-pi/2)^(pi/2) cos(u) sqrt(cos^2(u)) du
Simplify cos(u) sqrt(cos^2(u)) assuming -pi/2<u<pi/2:
  =  9 integral_(-pi/2)^(pi/2) cos^2(u) du
Write cos^2(u) as 1/2 cos(2 u)+1/2:
  =  9 integral_(-pi/2)^(pi/2) (1/2 cos(2 u)+1/2) du
Integrate the sum term by term and factor out constants:
  =  9/2 integral_(-pi/2)^(pi/2) cos(2 u) du+9/2 integral_(-pi/2)^(pi/2) 1 du
For the integrand cos(2 u), substitute s = 2 u and  ds = 2  du.
This gives a new lower bound s = (2 (-pi))/2 = -pi and upper bound s = (2 pi)/2 = pi:
  =  9/4 integral_(-pi)^pi cos(s) ds+9/2 integral_(-pi/2)^(pi/2) 1 du
Apply the fundamental theorem of calculus.
The antiderivative of cos(s) is sin(s):
  =  (9 sin(s))/4|_(-pi)^pi+9/2 integral_(-pi/2)^(pi/2) 1 du
Evaluate the antiderivative at the limits and subtract.
 (9 sin(s))/4|_(-pi)^pi = (9 sin(pi))/4-(9 sin(-pi))/4 = 0:
  =  9/2 integral_(-pi/2)^(pi/2) 1 du
Apply the fundamental theorem of calculus.
The antiderivative of 1 is u:
  =  (9 u)/2|_(-pi/2)^(pi/2)
Evaluate the antiderivative at the limits and subtract.
 (9 u)/2|_(-pi/2)^(pi/2) = (9 pi)/(2 2)-((9 (-pi))/(2 2)) = (9 pi)/2:
Answer: |  =  (9 pi)/2

It's the area of a semicircle of radius 3.  This is (1/2)*pi*3^2   or just 9pi/2.