What is the integral of sqrt(8^2-x^2) ?

The integral of sqrt(8^2-x^2) is 32(arcsin(1/8 x)+1/2 sin(2arcsin(1/8 x)))+C

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integral of sqrt(8^2-x^2)

\int 8^{2} - x^{2} d x

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

Solution steps

\int 8^{2} - x^{2} d x

Apply Trigonometric Substitution

: \int 64 cos^{2} (u) d u

= \int 64 cos^{2} (u) d u

Take the constant out:

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

= 64 \cdot \int cos^{2} (u) d u

Rewrite using trig identities

= 64 \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

= 64 \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

= 64 \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)

= 64 \cdot \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

u = arcsin (\frac{1}{8} x)

= 64 \cdot \frac{1}{2} (arcsin (\frac{1}{8} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{8} x)))

Simplify

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

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

Add a constant to the solution

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

\int_{- 1}^{1} (2 x^{2} - x^{3}) d x

\frac{d}{d x} (I n (x^{2} + 7))

\int \frac{x - 3}{( x + 8 ) ( x - 2 )} d x

a^{2} - x^{2}

h (x) = \frac{2 x}{sin ( x )}

What is the integral of sqrt(8^2-x^2) ?
The integral of sqrt(8^2-x^2) is 32(arcsin(1/8 x)+1/2 sin(2arcsin(1/8 x)))+C