What is the integral of sqrt(4/(9-9x^2)) ?

The integral of sqrt(4/(9-9x^2)) is 2/3 arcsin(x)+C

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integral of sqrt(4/(9-9x^2))

\int \frac{4}{9 - 9 x ^{2}} d x

\frac{2}{3} arcsin (x) + C

Solution steps

\int \frac{4}{9 - 9 x ^{2}} d x

Simplify

\frac{4}{9 - 9 x ^{2}} : \frac{2}{3 1 - x ^{2}}

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

Take the constant out:

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

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

Use the common integral:

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

= \frac{2}{3} arcsin (x)

Add a constant to the solution

= \frac{2}{3} arcsin (x) + C

x^{2} + 9 (- 5.34)

x \to 0 lim (3 x^{2} - 2 x + 4)

\frac{d}{d x} (15 x^{5} - \frac{60}{7 x ^{6}} - 1)

\int \frac{r}{( r ^{2} + z ^{2} ) ^{\frac{3}{2}}} d r

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

What is the integral of sqrt(4/(9-9x^2)) ?
The integral of sqrt(4/(9-9x^2)) is 2/3 arcsin(x)+C