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

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

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integral of (x^2)/(sqrt(4-x^2))

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int 4 sin^{2} (u) d u

= \int 4 sin^{2} (u) d u

Take the constant out:

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

= 4 \cdot \int sin^{2} (u) d u

Rewrite using trig identities

= 4 \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

= 4 \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

= 4 \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)

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

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

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

\frac{d x}{d t} = 2 x - x^{2}

\frac{\partial}{\partial x} (3 x - (\frac{3}{4} x + \frac{1}{4} y)^{2})

x \frac{d y}{d x} + (1 + x) y = e^{- x} sin (2 x)

\frac{d y}{d x} = - 1 + (2 x + y)^{2}

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