What is the integral of (1-x^2)^{-0.5} ?

The integral of (1-x^2)^{-0.5} is arcsin(x)+C

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integral of (1-x^2)^{-0.5}

\int (1 - x^{2})^{- 0.5} d x

arcsin (x) + C

Solution steps

\int (1 - x^{2})^{- 0.5} d x

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

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

Use the common integral:

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

= arcsin (x)

Add a constant to the solution

= arcsin (x) + C

e x p an d - \frac{z}{e ^{z + 1} ( z - 1 ) ^{2}}

s im pl i f y t^{- 1}

\frac{d}{d x} (x^{3} + 2)

x \to 0 lim (\frac{e ^{- 4 x} - 1}{x})

\int_{0}^{4} \frac{1}{16 + x ^{2}} d x

What is the integral of (1-x^2)^{-0.5} ?
The integral of (1-x^2)^{-0.5} is arcsin(x)+C