What is the integral of sqrt(4-3x^2) ?

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

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integral of sqrt(4-3x^2)

\int 4 - 3 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 - 3 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

\int \frac{e ^{2 x}}{16 - 9 e ^{4 x}} d x

\frac{d}{d x} (2 cos (x) - 4 x sin (x) - x^{2} cos (x))

\int \frac{1}{3 s h x + 5 c h x} d x

\frac{5 s}{( s ^{2} + 4 ) ( s + 1 )}

\frac{d y}{d x} = 0.002 y (500 - y) - 80

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