What is the integral of sqrt(36-4x^2) ?

The integral of sqrt(36-4x^2) is 9(arcsin(x/3)+1/2 sin(2arcsin(x/3)))+C

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integral of sqrt(36-4x^2)

\int 36 - 4 x^{2} d x

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

Solution steps

\int 36 - 4 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{2}{6} x)

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

Simplify

18 \cdot \frac{1}{2} (arcsin (\frac{2}{6} x) + \frac{1}{2} sin (2 arcsin (\frac{2}{6} x))) : 9 (arcsin (\frac{x}{3}) + \frac{1}{2} sin (2 arcsin (\frac{x}{3})))

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

Add a constant to the solution

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

\int 3 x e^{- 3 x} d x

y = x^{2} - 2 x, y = 3 x + 6

\int e^{- x} sin (4 x) d x

(x^{2} + 1) y^{^{'}} + 6 x (y - 1) = 0, y (0) = 8

\int (ln (y))^{2} d y

What is the integral of sqrt(36-4x^2) ?
The integral of sqrt(36-4x^2) is 9(arcsin(x/3)+1/2 sin(2arcsin(x/3)))+C