What is the integral of sqrt(-8+6x-x^2) ?

The integral of sqrt(-8+6x-x^2) is 1/2 (arcsin(x-3)+1/2 sin(2arcsin(x-3)))+C

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integral of sqrt(-8+6x-x^2)

\int - 8 + 6 x - x^{2} d x

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

Solution steps

\int - 8 + 6 x - x^{2} d x

Complete the square

- 8 + 6 x - x^{2} : - (x - 3)^{2} + 1

= \int - (x - 3)^{2} + 1 d x

Apply u-substitution

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

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

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

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

Substitute back

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

Add a constant to the solution

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

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What is the integral of sqrt(-8+6x-x^2) ?
The integral of sqrt(-8+6x-x^2) is 1/2 (arcsin(x-3)+1/2 sin(2arcsin(x-3)))+C