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

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

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integral of sqrt(x)sqrt(4-x)

\int x 4 - x d x

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

Solution steps

\int x 4 - x d x

Apply exponent rule:

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

x 4 - x = x (4 - x)

= \int x (4 - x) d x

Expand

x (4 - x) : 4 x - x^{2}

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

Complete the square

4 x - x^{2} : - (x - 2)^{2} + 4

= \int - (x - 2)^{2} + 4 d x

Apply u-substitution

: \int - u^{2} + 4 d u

= \int - u^{2} + 4 d u

Apply Trigonometric Substitution

: \int 4 cos^{2} (v) d v

= \int 4 cos^{2} (v) d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int cos^{2} (v) d v

Rewrite using trig identities

= 4 \cdot \int \frac{1 + cos ( 2 v )}{2} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \frac{1}{2} \cdot \int 1 + cos (2 v) d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 4 \cdot \frac{1}{2} (\int 1 d v + \int cos (2 v) d v)

\int 1 d v = v

\int cos (2 v) d v = \frac{1}{2} sin (2 v)

= 4 \cdot \frac{1}{2} (v + \frac{1}{2} sin (2 v))

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\int cos^{3} (x) sin (x) d x

y = \frac{a ^{4}}{a ^{3} - 8 x}

\frac{\partial}{\partial z} (z y^{2} (t + 1))

\int \frac{sec ( y )}{sin ( y )} d y

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