Question

How do you evaluate the integral `int sqrtxsqrt(4-x)`?

Answer 1

`4 sin^-1(sqrt x/2)-1/4sqrt x sqrt(4-x)(4-2x)+C`

Explanation:

Try `x = 4sin^2 theta`. Then

`sqrt x = 2 sin theta`
`sqrt(4-x) = 2 cos theta`
`dx = 8sin theta cos theta d theta`

Thus, the integral is

`int 2 sin theta times 2cos theta times 8 sin theta cos theta d theta = 32 int sin^2 theta cos^2 theta d theta`
`qquad = 8 int sin^2(2 theta)d theta = 4 int (1-cos (4 theta))d theta`
`qquad = 4 theta - sin (4 theta)+C = 4theta -4sin theta cos theta cos (2theta)+C`
`qquad = 4 sin^-1(sqrt x/2)-1/4sqrt x sqrt(4-x)(4-2x)+C`