What is the integral of 1/(sqrt(4-4x^2)) ?

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

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integral of 1/(sqrt(4-4x^2))

\int \frac{1}{4 - 4 x ^{2}} d x

\frac{1}{2} arcsin (x) + C

Solution steps

\int \frac{1}{4 - 4 x ^{2}} d x

4 - 4 x^{2} = 2 1 - x^{2}

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

Take the constant out:

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

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

Use the common integral:

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

= \frac{1}{2} arcsin (x)

Add a constant to the solution

= \frac{1}{2} arcsin (x) + C

\frac{1}{1 + x}

\int \frac{13}{x 1 + 2 x} d x

(30 + x) (230 - 10 x)

\frac{d}{d x} (\frac{5 x}{e ^{x}})

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

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