What is the integral of sqrt(4-9x^2) ?

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

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integral of sqrt(4-9x^2)

\int 4 - 9 x^{2} d x

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

Solution steps

\int 4 - 9 x^{2} d x

Apply Trigonometric Substitution

: \int \frac{4}{3} cos^{2} (u) d u

= \int \frac{4}{3} cos^{2} (u) d u

Take the constant out:

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

= \frac{4}{3} \cdot \int cos^{2} (u) d u

Rewrite using trig identities

= \frac{4}{3} \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

= \frac{4}{3} \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

= \frac{4}{3} \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)

= \frac{4}{3} \cdot \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

e^{x - y}

y = 7

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

c^{\frac{1}{2}}

\frac{- 2}{x} - \frac{- 4}{y} = 1

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