What is the integral of sqrt(10x-x^2) ?

The integral of sqrt(10x-x^2) is 25/4 (2arcsin(1/5 x-1)+sin(2arcsin(1/5 x-1)))+C

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integral of sqrt(10x-x^2)

\int 10 x - x^{2} d x

\frac{25}{4} (2 arcsin (\frac{1}{5} x - 1) + sin (2 arcsin (\frac{1}{5} x - 1))) + C

Solution steps

\int 10 x - x^{2} d x

Complete the square

10 x - x^{2} : - (x - 5)^{2} + 25

= \int - (x - 5)^{2} + 25 d x

Apply u-substitution

: \int - u^{2} + 25 d u

= \int - u^{2} + 25 d u

Apply Trigonometric Substitution

: \int 25 cos^{2} (v) d v

= \int 25 cos^{2} (v) d v

Take the constant out:

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

= 25 \cdot \int cos^{2} (v) d v

Rewrite using trig identities

= 25 \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

= 25 \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

= 25 \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)

= 25 \cdot \frac{1}{2} (v + \frac{1}{2} sin (2 v))

Substitute back

= 25 \cdot \frac{1}{2} (arcsin (\frac{1}{5} (x - 5)) + \frac{1}{2} sin (2 arcsin (\frac{1}{5} (x - 5))))

Simplify

25 \cdot \frac{1}{2} (arcsin (\frac{1}{5} (x - 5)) + \frac{1}{2} sin (2 arcsin (\frac{1}{5} (x - 5)))) : \frac{25}{4} (2 arcsin (\frac{1}{5} x - 1) + sin (2 arcsin (\frac{1}{5} x - 1)))

= \frac{25}{4} (2 arcsin (\frac{1}{5} x - 1) + sin (2 arcsin (\frac{1}{5} x - 1)))

Add a constant to the solution

= \frac{25}{4} (2 arcsin (\frac{1}{5} x - 1) + sin (2 arcsin (\frac{1}{5} x - 1))) + C

\frac{3}{( s ^{2} + 9 )}

\frac{1}{x}, - 1

\int \frac{3 x ^{5}}{5 - 12 x ^{6}} d x

\int_{1}^{e} \frac{2}{x} d x

\int (- 1)^{n}

What is the integral of sqrt(10x-x^2) ?
The integral of sqrt(10x-x^2) is 25/4 (2arcsin(1/5 x-1)+sin(2arcsin(1/5 x-1)))+C