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

- \frac{z}{e ^{z + 1} ( z - 1 ) ^{2}}

\frac{d}{d x} (x^{2} + 2 x - x^{- 1} + x^{- 3})

h^{^{'}} = 2 t^{2} - 3 e^{t}

f (x) = \frac{2}{x}, at x = 3

t^{- 1}

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