What is the integral of sqrt(64-4x^2) ?

The integral of sqrt(64-4x^2) is 16(arcsin(x/4)+1/2 sin(2arcsin(x/4)))+C

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integral of sqrt(64-4x^2)

\int 64 - 4 x^{2} d x

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

Solution steps

\int 64 - 4 x^{2} d x

Apply Trigonometric Substitution

: \int 32 cos^{2} (u) d u

= \int 32 cos^{2} (u) d u

Take the constant out:

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

= 32 \cdot \int cos^{2} (u) d u

Rewrite using trig identities

= 32 \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

= 32 \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

= 32 \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)

= 32 \cdot \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

u = arcsin (\frac{2}{8} x)

= 32 \cdot \frac{1}{2} (arcsin (\frac{2}{8} x) + \frac{1}{2} sin (2 arcsin (\frac{2}{8} x)))

Simplify

32 \cdot \frac{1}{2} (arcsin (\frac{2}{8} x) + \frac{1}{2} sin (2 arcsin (\frac{2}{8} x))) : 16 (arcsin (\frac{x}{4}) + \frac{1}{2} sin (2 arcsin (\frac{x}{4})))

= 16 (arcsin (\frac{x}{4}) + \frac{1}{2} sin (2 arcsin (\frac{x}{4})))

Add a constant to the solution

= 16 (arcsin (\frac{x}{4}) + \frac{1}{2} sin (2 arcsin (\frac{x}{4}))) + C

\int 3 w 4 - w^{2} d w

\frac{\partial}{\partial x} (e^{y} + x + x^{4} + 5)

5 + 2 x, \frac{( 15 + x )}{3}

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

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

What is the integral of sqrt(64-4x^2) ?
The integral of sqrt(64-4x^2) is 16(arcsin(x/4)+1/2 sin(2arcsin(x/4)))+C