What is the integral of sqrt(x(4-x)) ?

The integral of sqrt(x(4-x)) is -4arcsin(1/2 sqrt(4-x))+sin(4arcsin(1/2 sqrt(4-x)))+C

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integral of sqrt(x(4-x))

\int x (4 - x) d x

- 4 arcsin (\frac{1}{2} 4 - x) + sin (4 arcsin (\frac{1}{2} 4 - x)) + C

Solution steps

\int x (4 - x) d x

Apply exponent rule:

(a \cdot b)^{n} = a^{n} \cdot b^{n}

x (4 - x) = x (4 - x)

= \int x 4 - x d x

Apply Integration By Parts

: \frac{2}{3} x^{\frac{3}{2}} 4 - x - \int - \frac{x ^{\frac{3}{2}}}{3 4 - x} d x

= \frac{2}{3} x^{\frac{3}{2}} 4 - x - \int - \frac{x ^{\frac{3}{2}}}{3 4 - x} d x

\int - \frac{x ^{\frac{3}{2}}}{3 4 - x} d x = \frac{2}{3} (x^{\frac{3}{2}} 4 - x + 6 (arcsin (\frac{1}{2} 4 - x) - \frac{1}{4} sin (4 arcsin (\frac{1}{2} 4 - x))))

= \frac{2}{3} x^{\frac{3}{2}} 4 - x - \frac{2}{3} (x^{\frac{3}{2}} 4 - x + 6 (arcsin (\frac{1}{2} 4 - x) - \frac{1}{4} sin (4 arcsin (\frac{1}{2} 4 - x))))

Simplify

\frac{2}{3} x^{\frac{3}{2}} 4 - x - \frac{2}{3} (x^{\frac{3}{2}} 4 - x + 6 (arcsin (\frac{1}{2} 4 - x) - \frac{1}{4} sin (4 arcsin (\frac{1}{2} 4 - x)))) : - 4 arcsin (\frac{1}{2} 4 - x) + sin (4 arcsin (\frac{1}{2} 4 - x))

= - 4 arcsin (\frac{1}{2} 4 - x) + sin (4 arcsin (\frac{1}{2} 4 - x))

Add a constant to the solution

= - 4 arcsin (\frac{1}{2} 4 - x) + sin (4 arcsin (\frac{1}{2} 4 - x)) + C

f (x) = (\frac{x}{1 + x})^{5}, at x = 1

8^{t} \cdot t + 10

\frac{d}{d y} (tan (x^{2} + y))

x^{3} - 4

\int e^{2 x y} d x

What is the integral of sqrt(x(4-x)) ?
The integral of sqrt(x(4-x)) is -4arcsin(1/2 sqrt(4-x))+sin(4arcsin(1/2 sqrt(4-x)))+C