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

May 31, 2018

$4 {\sin}^{-} 1 \left(\frac{\sqrt{x}}{2}\right) - \frac{1}{4} \sqrt{x} \sqrt{4 - x} \left(4 - 2 x\right) + C$

#### Explanation:

Try $x = 4 {\sin}^{2} \theta$. Then

$\sqrt{x} = 2 \sin \theta$
$\sqrt{4 - x} = 2 \cos \theta$
$\mathrm{dx} = 8 \sin \theta \cos \theta d \theta$

Thus, the integral is

$\int 2 \sin \theta \times 2 \cos \theta \times 8 \sin \theta \cos \theta d \theta = 32 \int {\sin}^{2} \theta {\cos}^{2} \theta d \theta$
$q \quad = 8 \int {\sin}^{2} \left(2 \theta\right) d \theta = 4 \int \left(1 - \cos \left(4 \theta\right)\right) d \theta$
$q \quad = 4 \theta - \sin \left(4 \theta\right) + C = 4 \theta - 4 \sin \theta \cos \theta \cos \left(2 \theta\right) + C$
$q \quad = 4 {\sin}^{-} 1 \left(\frac{\sqrt{x}}{2}\right) - \frac{1}{4} \sqrt{x} \sqrt{4 - x} \left(4 - 2 x\right) + C$