What is the integral of sqrt(16-4x^2) ?

The integral of sqrt(16-4x^2) is 4(arcsin(x/2)+1/2 sin(2arcsin(x/2)))+C

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integral of sqrt(16-4x^2)

\int 16 - 4 x^{2} d x

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

Solution steps

\int 16 - 4 x^{2} d x

Apply Trigonometric Substitution

: \int 8 cos^{2} (u) d u

= \int 8 cos^{2} (u) d u

Take the constant out:

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

= 8 \cdot \int cos^{2} (u) d u

Rewrite using trig identities

= 8 \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

= 8 \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

= 8 \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)

= 8 \cdot \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

u = arcsin (\frac{2}{4} x)

= 8 \cdot \frac{1}{2} (arcsin (\frac{2}{4} x) + \frac{1}{2} sin (2 arcsin (\frac{2}{4} x)))

Simplify

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

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

Add a constant to the solution

= 4 (arcsin (\frac{x}{2}) + \frac{1}{2} sin (2 arcsin (\frac{x}{2}))) + C

\frac{1}{2} sec (2 x)

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

x \to 1 lim (4 x^{2} - 2 x + 5)

\frac{d}{d x} (\frac{1 + \frac{1}{x}}{1 - \frac{1}{x}})

\frac{d y}{d x} = e^{3 x} \cdot e^{2 y}

What is the integral of sqrt(16-4x^2) ?
The integral of sqrt(16-4x^2) is 4(arcsin(x/2)+1/2 sin(2arcsin(x/2)))+C