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

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

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

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

arcsin (x) + C

Solution steps

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

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 (x)

= 1 \cdot arcsin (x)

Simplify

= arcsin (x)

Add a constant to the solution

= arcsin (x) + C

\frac{d}{d x} (100 x^{\frac{3}{2}})

n = 1 \sum \infty (\frac{4}{9})^{2 n + 1}

f (x) = x^{2} + x, at x = 2

y^{^{''}} + 3 y^{^{'}} + 3 y = 0

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

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