What is the integral of 1/(sqrt(8x-x^2)) ?

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

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integral of 1/(sqrt(8x-x^2))

\int \frac{1}{8 x - x ^{2}} d x

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

Solution steps

\int \frac{1}{8 x - x ^{2}} d x

Complete the square

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

= \int \frac{1}{- ( x - 4 ) ^{2} + 16} d x

Apply u-substitution

: \int \frac{1}{- u ^{2} + 16} d u

= \int \frac{1}{- u ^{2} + 16} d u

Apply Trigonometric Substitution

: \int 1 d v

= \int 1 d v

Integral of a constant:

\int a d x = a x

= 1 \cdot v

Substitute back

= 1 \cdot arcsin (\frac{1}{4} (x - 4))

Simplify

1 \cdot arcsin (\frac{1}{4} (x - 4)) : arcsin (\frac{1}{4} x - 1)

= arcsin (\frac{1}{4} x - 1)

Add a constant to the solution

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

5 3 x

\frac{\partial}{\partial y} (x ln (yz))

\int x^{4} + 10 x^{2} + 25 d x

y^{^{'}} + \frac{1}{2} y = 17 sin (2 t)

y = 4^{9^{x^{2}}}

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