What is the integral of sqrt(256-x^2) ?

The integral of sqrt(256-x^2) is 128(arcsin(1/16 x)+1/2 sin(2arcsin(1/16 x)))+C

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integral of sqrt(256-x^2)

\int 256 - x^{2} d x

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

Solution steps

\int 256 - x^{2} d x

Apply Trigonometric Substitution

: \int 256 cos^{2} (u) d u

= \int 256 cos^{2} (u) d u

Take the constant out:

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

= 256 \cdot \int cos^{2} (u) d u

Rewrite using trig identities

= 256 \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

= 256 \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

= 256 \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)

= 256 \cdot \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

u = arcsin (\frac{1}{16} x)

= 256 \cdot \frac{1}{2} (arcsin (\frac{1}{16} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{16} x)))

Simplify

256 \cdot \frac{1}{2} (arcsin (\frac{1}{16} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{16} x))) : 128 (arcsin (\frac{1}{16} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{16} x)))

= 128 (arcsin (\frac{1}{16} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{16} x)))

Add a constant to the solution

= 128 (arcsin (\frac{1}{16} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{16} x))) + C

\int \frac{2 x}{2 x ^{2} + 1} d x

\frac{d y}{d x} = \frac{3 x ^{2}}{e ^{2 y}}, y (0) = \frac{1}{2}

2 x (x^{2} - 4 x + 5)^{8}, at x = 2

y^{^{''}} + 3 y^{^{'}} + 2 y = 12 t^{3}

(\frac{2}{x ^{3}})^{^{'}}

What is the integral of sqrt(256-x^2) ?
The integral of sqrt(256-x^2) is 128(arcsin(1/16 x)+1/2 sin(2arcsin(1/16 x)))+C