What is the integral of sqrt(-x^2+4x) ?

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

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integral of sqrt(-x^2+4x)

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

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

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

Apply Trigonometric Substitution

= \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

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