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

The integral of sqrt(8x-x^2) is 8(arcsin(1/4 x-1)+1/2 sin(2arcsin(1/4 x-1)))+C

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integral of sqrt(8x-x^2)

integral

8 x - x^{2}

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

Solution steps

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

Complete the square

8 x - x^{2} : - (x - 4)^{2} + 16

= \int - (x - 4)^{2} + 16 d x

Apply u-substitution

: \int - u^{2} + 16 d u

= \int - u^{2} + 16 d u

Apply Trigonometric Substitution

: \int 16 cos^{2} (v) d v

= \int 16 cos^{2} (v) d v

Take the constant out:

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

= 16 \cdot \int cos^{2} (v) d v

Rewrite using trig identities

= 16 \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

= 16 \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

= 16 \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)

= 16 \cdot \frac{1}{2} (v + \frac{1}{2} sin (2 v))

Substitute back

= 16 \cdot \frac{1}{2} (arcsin (\frac{1}{4} (x - 4)) + \frac{1}{2} sin (2 arcsin (\frac{1}{4} (x - 4))))

Simplify

16 \cdot \frac{1}{2} (arcsin (\frac{1}{4} (x - 4)) + \frac{1}{2} sin (2 arcsin (\frac{1}{4} (x - 4)))) : 8 (arcsin (\frac{1}{4} x - 1) + \frac{1}{2} sin (2 arcsin (\frac{1}{4} x - 1)))

= 8 (arcsin (\frac{1}{4} x - 1) + \frac{1}{2} sin (2 arcsin (\frac{1}{4} x - 1)))

Add a constant to the solution

= 8 (arcsin (\frac{1}{4} x - 1) + \frac{1}{2} sin (2 arcsin (\frac{1}{4} x - 1))) + C

(\frac{x ^{2} - 1}{x})^{3}

\frac{( x ^{24} )}{( x ^{25} )}

\frac{( 2 v + 5 )}{( 7 v + 22 )}

\frac{x ^{2}}{( a ^{2} + x ^{2} ) ^{2}}

\frac{x ^{3}}{x - 2}

What is the integral of sqrt(8x-x^2) ?
The integral of sqrt(8x-x^2) is 8(arcsin(1/4 x-1)+1/2 sin(2arcsin(1/4 x-1)))+C