What is the integral of sqrt(18-x^2) ?

The integral of sqrt(18-x^2) is 9(arcsin(1/(3sqrt(2))x)+1/2 sin(2arcsin(1/(3sqrt(2))x)))+C

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integral of sqrt(18-x^2)

\int 18 - x^{2} d x

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

Solution steps

\int 18 - x^{2} d x

Apply Trigonometric Substitution

: \int 18 cos^{2} (u) d u

= \int 18 cos^{2} (u) d u

Take the constant out:

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

= 18 \cdot \int cos^{2} (u) d u

Rewrite using trig identities

= 18 \cdot \int \frac{1 + cos ( 2 u )}{2} d u

Take the constant out:

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

= 18 \cdot \frac{1}{2} \cdot \int 1 + cos (2 u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 18 \cdot \frac{1}{2} (\int 1 d u + \int cos (2 u) d u)

\int 1 d u = u

\int cos (2 u) d u = \frac{1}{2} sin (2 u)

= 18 \cdot \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

u = arcsin (\frac{1}{3 2} x)

= 18 \cdot \frac{1}{2} (arcsin (\frac{1}{3 2} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{3 2} x)))

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (arccot (3 x))

x^{\frac{5}{3}} - x^{\frac{2}{3}}

x - 3

f (t) = t^{\frac{2}{3}} - t^{\frac{1}{3}} + 4

x \to 3 lim (5 x - 3)

What is the integral of sqrt(18-x^2) ?
The integral of sqrt(18-x^2) is 9(arcsin(1/(3sqrt(2))x)+1/2 sin(2arcsin(1/(3sqrt(2))x)))+C