What is the integral of (10)/(sqrt(8x-4x^2)) ?

The integral of (10)/(sqrt(8x-4x^2)) is 5arcsin(x-1)+C

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integral of (10)/(sqrt(8x-4x^2))

\int \frac{10}{8 x - 4 x ^{2}} d x

5 arcsin (x - 1) + C

Solution steps

\int \frac{10}{8 x - 4 x ^{2}} d x

Take the constant out:

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

= 10 \cdot \int \frac{1}{8 x - 4 x ^{2}} d x

8 x - 4 x^{2} = 2 x (2 - x)

= 10 \cdot \int \frac{1}{2 x ( 2 - x )} d x

Take the constant out:

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

= 10 \cdot \frac{1}{2} \cdot \int \frac{1}{x ( 2 - x )} d x

Complete the square

x (2 - x) : - (x - 1)^{2} + 1

= 10 \cdot \frac{1}{2} \cdot \int \frac{1}{- ( x - 1 ) ^{2} + 1} d x

Apply u-substitution

= 10 \cdot \frac{1}{2} \cdot \int \frac{1}{- u ^{2} + 1} d u

Use the common integral:

\int \frac{1}{- u ^{2} + 1} d u = arcsin (u)

= 10 \cdot \frac{1}{2} arcsin (u)

Substitute back

u = x - 1

= 10 \cdot \frac{1}{2} arcsin (x - 1)

Simplify

10 \cdot \frac{1}{2} arcsin (x - 1) : 5 arcsin (x - 1)

= 5 arcsin (x - 1)

Add a constant to the solution

= 5 arcsin (x - 1) + C

What is the integral of (10)/(sqrt(8x-4x^2)) ?
The integral of (10)/(sqrt(8x-4x^2)) is 5arcsin(x-1)+C