What is the integral of 8/(sqrt(9-9x^2)) ?

The integral of 8/(sqrt(9-9x^2)) is 8/3 arcsin(x)+C

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integral of 8/(sqrt(9-9x^2))

\int \frac{8}{9 - 9 x ^{2}} d x

\frac{8}{3} arcsin (x) + C

Solution steps

\int \frac{8}{9 - 9 x ^{2}} d x

Take the constant out:

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

= 8 \cdot \int \frac{1}{9 - 9 x ^{2}} d x

9 - 9 x^{2} = 3 1 - x^{2}

= 8 \cdot \int \frac{1}{3 1 - x ^{2}} d x

Take the constant out:

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

= 8 \cdot \frac{1}{3} \cdot \int \frac{1}{1 - x ^{2}} d x

Use the common integral:

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

= 8 \cdot \frac{1}{3} arcsin (x)

Simplify

8 \cdot \frac{1}{3} arcsin (x) : \frac{8}{3} arcsin (x)

= \frac{8}{3} arcsin (x)

Add a constant to the solution

= \frac{8}{3} arcsin (x) + C

What is the integral of 8/(sqrt(9-9x^2)) ?
The integral of 8/(sqrt(9-9x^2)) is 8/3 arcsin(x)+C