What is the integral of sqrt(20x-4x^2-24) ?

The integral of sqrt(20x-4x^2-24) is 1/8 (2arcsin(2x-5)+sin(2arcsin(2x-5)))+C

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integral of sqrt(20x-4x^2-24)

\int 20 x - 4 x^{2} - 24 d x

\frac{1}{8} (2 arcsin (2 x - 5) + sin (2 arcsin (2 x - 5))) + C

Solution steps

\int 20 x - 4 x^{2} - 24 d x

Complete the square

20 x - 4 x^{2} - 24 : - 4 (x - \frac{5}{2})^{2} + 1

= \int - 4 (x - \frac{5}{2})^{2} + 1 d x

Apply u-substitution

: \int - 4 u^{2} + 1 d u

= \int - 4 u^{2} + 1 d u

Apply Trigonometric Substitution

: \int \frac{1}{2} cos^{2} (v) d v

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

Take the constant out:

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

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

Rewrite using trig identities

= \frac{1}{2} \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

= \frac{1}{2} \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

= \frac{1}{2} \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)

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

Substitute back

= \frac{1}{2} \cdot \frac{1}{2} (arcsin (2 (x - \frac{5}{2})) + \frac{1}{2} sin (2 arcsin (2 (x - \frac{5}{2}))))

Simplify

\frac{1}{2} \cdot \frac{1}{2} (arcsin (2 (x - \frac{5}{2})) + \frac{1}{2} sin (2 arcsin (2 (x - \frac{5}{2})))) : \frac{1}{8} (2 arcsin (2 x - 5) + sin (2 arcsin (2 x - 5)))

= \frac{1}{8} (2 arcsin (2 x - 5) + sin (2 arcsin (2 x - 5)))

Add a constant to the solution

= \frac{1}{8} (2 arcsin (2 x - 5) + sin (2 arcsin (2 x - 5))) + C

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What is the integral of sqrt(20x-4x^2-24) ?
The integral of sqrt(20x-4x^2-24) is 1/8 (2arcsin(2x-5)+sin(2arcsin(2x-5)))+C