What is the integral of sqrt(16x-x^2) ?

The integral of sqrt(16x-x^2) is 32(arcsin(1/8 x-1)+1/2 sin(2arcsin(1/8 x-1)))+C

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integral of sqrt(16x-x^2)

\int 16 x - x^{2} d x

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

Solution steps

\int 16 x - x^{2} d x

Complete the square

16 x - x^{2} : - (x - 8)^{2} + 64

= \int - (x - 8)^{2} + 64 d x

Apply u-substitution

: \int - u^{2} + 64 d u

= \int - u^{2} + 64 d u

Apply Trigonometric Substitution

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

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

Take the constant out:

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

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

Rewrite using trig identities

= 64 \cdot \int \frac{1 + cos ( 2 v )}{2} d v

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 v) d v

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 v + \int cos (2 v) d v)

\int 1 d v = v

\int cos (2 v) d v = \frac{1}{2} sin (2 v)

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of sqrt(16x-x^2) ?
The integral of sqrt(16x-x^2) is 32(arcsin(1/8 x-1)+1/2 sin(2arcsin(1/8 x-1)))+C