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

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

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integral of sqrt(4-4x^2)

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

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

Solution steps

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

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

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

Simplify

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

= \int 2 1 - x^{2} d x

Take the constant out:

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

= 2 \cdot \int 1 - x^{2} d x

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

= 2 \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

= 2 \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

= 2 \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)

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

Substitute back

u = arcsin (x)

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

Simplify

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

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

Add a constant to the solution

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

2 x^{4}

\frac{d}{d x} (\frac{x ^{3}}{8})

x \to - 4 + lim (\frac{2}{( x + 4 ) ^{3}})

\int \frac{x ^{2} - 49}{x ^{2}} d x

y^{^{''}} + \frac{2}{t} y^{^{'}} = 0

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