What is the integral of sqrt(6x-x^2) ?

The integral of sqrt(6x-x^2) is 9/4 (2arcsin(1/3 x-1)+sin(2arcsin(1/3 x-1)))+C

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integral of sqrt(6x-x^2)

\int 6 x - x^{2} d x

\frac{9}{4} (2 arcsin (\frac{1}{3} x - 1) + sin (2 arcsin (\frac{1}{3} x - 1))) + C

Solution steps

\int 6 x - x^{2} d x

Complete the square

6 x - x^{2} : - (x - 3)^{2} + 9

= \int - (x - 3)^{2} + 9 d x

Apply u-substitution

: \int - u^{2} + 9 d u

= \int - u^{2} + 9 d u

Apply Trigonometric Substitution

: \int 9 cos^{2} (v) d v

= \int 9 cos^{2} (v) d v

Take the constant out:

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

= 9 \cdot \int cos^{2} (v) d v

Rewrite using trig identities

= 9 \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

= 9 \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

= 9 \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)

= 9 \cdot \frac{1}{2} (v + \frac{1}{2} sin (2 v))

Substitute back

= 9 \cdot \frac{1}{2} (arcsin (\frac{1}{3} (x - 3)) + \frac{1}{2} sin (2 arcsin (\frac{1}{3} (x - 3))))

Simplify

9 \cdot \frac{1}{2} (arcsin (\frac{1}{3} (x - 3)) + \frac{1}{2} sin (2 arcsin (\frac{1}{3} (x - 3)))) : \frac{9}{4} (2 arcsin (\frac{1}{3} x - 1) + sin (2 arcsin (\frac{1}{3} x - 1)))

= \frac{9}{4} (2 arcsin (\frac{1}{3} x - 1) + sin (2 arcsin (\frac{1}{3} x - 1)))

Add a constant to the solution

= \frac{9}{4} (2 arcsin (\frac{1}{3} x - 1) + sin (2 arcsin (\frac{1}{3} x - 1))) + C

f (x) = (2 x^{3} + 4 x + 1)

\int (4 + 4 tan^{2} (v)) d v

(sin (\frac{x}{2}))^{^{'}}

\frac{\partial}{\partial x} (x y^{2} + y e^{x^{2}} + 5)

\frac{d y}{d x} + 2 x y = 2 x^{3}, y (0) = 2

What is the integral of sqrt(6x-x^2) ?
The integral of sqrt(6x-x^2) is 9/4 (2arcsin(1/3 x-1)+sin(2arcsin(1/3 x-1)))+C