What is the integral of sqrt(12x-x^2) ?

The integral of sqrt(12x-x^2) is 18(arcsin(1/6 x-1)+1/2 sin(2arcsin(1/6 x-1)))+C

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integral of sqrt(12x-x^2)

\int 12 x - x^{2} d x

18 (arcsin (\frac{1}{6} x - 1) + \frac{1}{2} sin (2 arcsin (\frac{1}{6} x - 1))) + C

Solution steps

\int 12 x - x^{2} d x

Complete the square

12 x - x^{2} : - (x - 6)^{2} + 36

= \int - (x - 6)^{2} + 36 d x

Apply u-substitution

: \int - u^{2} + 36 d u

= \int - u^{2} + 36 d u

Apply Trigonometric Substitution

: \int 36 cos^{2} (v) d v

= \int 36 cos^{2} (v) d v

Take the constant out:

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

= 36 \cdot \int cos^{2} (v) d v

Rewrite using trig identities

= 36 \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

= 36 \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

= 36 \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)

= 36 \cdot \frac{1}{2} (v + \frac{1}{2} sin (2 v))

Substitute back

= 36 \cdot \frac{1}{2} (arcsin (\frac{1}{6} (x - 6)) + \frac{1}{2} sin (2 arcsin (\frac{1}{6} (x - 6))))

Simplify

36 \cdot \frac{1}{2} (arcsin (\frac{1}{6} (x - 6)) + \frac{1}{2} sin (2 arcsin (\frac{1}{6} (x - 6)))) : 18 (arcsin (\frac{1}{6} x - 1) + \frac{1}{2} sin (2 arcsin (\frac{1}{6} x - 1)))

= 18 (arcsin (\frac{1}{6} x - 1) + \frac{1}{2} sin (2 arcsin (\frac{1}{6} x - 1)))

Add a constant to the solution

= 18 (arcsin (\frac{1}{6} x - 1) + \frac{1}{2} sin (2 arcsin (\frac{1}{6} x - 1))) + C

y = 4 x, y = 9 x^{2}

\frac{d}{d x} (ln (sech (2 x)))

\int_{\frac{π}{6}}^{π} (sin (x)) d x

\frac{d}{d x} (\frac{x ^{2}}{ln ( x )} + 2 x)

What is the integral of sqrt(12x-x^2) ?
The integral of sqrt(12x-x^2) is 18(arcsin(1/6 x-1)+1/2 sin(2arcsin(1/6 x-1)))+C