What is the integral of (x^2)/((4-x^2)^{1/2)} ?

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

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integral of (x^2)/((4-x^2)^{1/2)}

\int \frac{x ^{2}}{( 4 - x ^{2} ) ^{\frac{1}{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} ) ^{\frac{1}{2}}} d x

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{x ^{2}}{( 4 - x ^{2} ) ^{\frac{1}{2}}} = (x^{2}) (4 - x^{2})^{- \frac{1}{2}}

= \int x^{2} (4 - x^{2})^{- \frac{1}{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

\frac{d}{d x} (x^{2} + x^{4})

f (x) = \frac{1}{1 - x ^{2}}

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

f (x) = (1 - x) (x^{2} - 8)^{2}, at x = 3

f (x) = 7 arcsin (x^{4})

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