What is the integral of (1-y^2)^{-1/2} ?

The integral of (1-y^2)^{-1/2} is arcsin(y)+C

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integral of (1-y^2)^{-1/2}

\int (1 - y^{2})^{- \frac{1}{2}} d y

arcsin (y) + C

Solution steps

\int (1 - y^{2})^{- \frac{1}{2}} d y

Apply Trigonometric Substitution

: \int 1 d u

= \int 1 d u

Integral of a constant:

\int a d x = a x

= 1 \cdot u

Substitute back

u = arcsin (y)

= 1 \cdot arcsin (y)

Simplify

= arcsin (y)

Add a constant to the solution

= arcsin (y) + C

y^{^{'}} + \frac{y}{x} = 9 x^{3} y^{2}

\frac{d}{d x} (n^{x})

y^{^{''}} + 6 y^{^{'}} + 9 y = 0

\frac{d y}{d t} - (\frac{3}{t}) y = 2 t^{3} e^{2 t}, y (1) = 0

\int \frac{x + 5}{x - 2} d x

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