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{d}{d x} (2 - sin (x))

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

\int_{0}^{2} 7 x^{3} x^{2} + 4 d x

\frac{d}{d x} (\frac{x ^{2} + 6 x}{10 - x ^{2}})

y^{^{''}} - 7 y^{^{'}} + 10 y = - sin (3 t)

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