What is the integral of sqrt(1-64x^2) ?

The integral of sqrt(1-64x^2) is 1/16 (arcsin(8x)+1/2 sin(2arcsin(8x)))+C

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integral of sqrt(1-64x^2)

\int 1 - 64 x^{2} d x

\frac{1}{16} (arcsin (8 x) + \frac{1}{2} sin (2 arcsin (8 x))) + C

Solution steps

\int 1 - 64 x^{2} d x

Apply Trigonometric Substitution

: \int \frac{1}{8} cos^{2} (u) d u

= \int \frac{1}{8} cos^{2} (u) d u

Take the constant out:

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

= \frac{1}{8} \cdot \int cos^{2} (u) d u

Rewrite using trig identities

= \frac{1}{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

= \frac{1}{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

= \frac{1}{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)

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

Substitute back

u = arcsin (8 x)

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

Simplify

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

= \frac{1}{16} (arcsin (8 x) + \frac{1}{2} sin (2 arcsin (8 x)))

Add a constant to the solution

= \frac{1}{16} (arcsin (8 x) + \frac{1}{2} sin (2 arcsin (8 x))) + C

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

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

\frac{d}{d t} (2)

\int x + x e^{x} d x

x^{3} + x^{2} - 6 x, 6 x

What is the integral of sqrt(1-64x^2) ?
The integral of sqrt(1-64x^2) is 1/16 (arcsin(8x)+1/2 sin(2arcsin(8x)))+C