How do you evaluate the integral int sqrtxsqrt(4-x)∫√x√4−x?

1 Answer
Ananda Dasgupta
May 31, 2018

4 sin^-1(sqrt x/2)-1/4sqrt x sqrt(4-x)(4-2x)+C4sin−1(√x2)−14√x√4−x(4−2x)+C

Explanation:

Try x = 4sin^2 thetax=4sin2θ. Then

sqrt x = 2 sin theta√x=2sinθ
sqrt(4-x) = 2 cos theta√4−x=2cosθ
dx = 8sin theta cos theta d theta dx=8sinθcosθdθ

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∫2sinθ×2cosθ×8sinθcosθdθ=32∫sin2θcos2θdθ
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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