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

1 Answer
Ananda Dasgupta
May 31, 2018

#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#

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