What is the integral of sqrt(14x-x^2) ?

The integral of sqrt(14x-x^2) is 49/4 (2arcsin(1/7 x-1)+sin(2arcsin(1/7 x-1)))+C

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integral of sqrt(14x-x^2)

\int 14 x - x^{2} d x

\frac{49}{4} (2 arcsin (\frac{1}{7} x - 1) + sin (2 arcsin (\frac{1}{7} x - 1))) + C

Solution steps

\int 14 x - x^{2} d x

Complete the square

14 x - x^{2} : - (x - 7)^{2} + 49

= \int - (x - 7)^{2} + 49 d x

Apply u-substitution

= \int - u^{2} + 49 d u

Apply Trigonometric Substitution

= \int 49 cos^{2} (v) d v

Take the constant out:

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

= 49 \cdot \int cos^{2} (v) d v

Rewrite using trig identities

= 49 \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

= 49 \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

= 49 \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)

= 49 \cdot \frac{1}{2} (v + \frac{1}{2} sin (2 v))

Substitute back

= 49 \cdot \frac{1}{2} (arcsin (\frac{1}{7} (x - 7)) + \frac{1}{2} sin (2 arcsin (\frac{1}{7} (x - 7))))

Simplify

49 \cdot \frac{1}{2} (arcsin (\frac{1}{7} (x - 7)) + \frac{1}{2} sin (2 arcsin (\frac{1}{7} (x - 7)))) : \frac{49}{4} (2 arcsin (\frac{1}{7} x - 1) + sin (2 arcsin (\frac{1}{7} x - 1)))

= \frac{49}{4} (2 arcsin (\frac{1}{7} x - 1) + sin (2 arcsin (\frac{1}{7} x - 1)))

Add a constant to the solution

= \frac{49}{4} (2 arcsin (\frac{1}{7} x - 1) + sin (2 arcsin (\frac{1}{7} x - 1))) + C

\frac{d}{d x} (x^{\frac{- 2}{5}})

f (x) = 2 x cos (x^{2})

\int_{0}^{2} \frac{7}{x - 2} d x

k = 1 \sum \infty ln (k)

t an g e n t f (x) = arcsin (3 x), at x = \frac{1}{6}

What is the integral of sqrt(14x-x^2) ?
The integral of sqrt(14x-x^2) is 49/4 (2arcsin(1/7 x-1)+sin(2arcsin(1/7 x-1)))+C