What is the integral from 0 to x of (1-t^2)^{-1/2} ?

The integral from 0 to x of (1-t^2)^{-1/2} is arcsin(x)

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integral from 0 to x of (1-t^2)^{-1/2}

\int_{0}^{x} (1 - t^{2})^{- \frac{1}{2}} d t

arcsin (x)

Solution steps

\int_{0}^{x} (1 - t^{2})^{- \frac{1}{2}} d t

Apply Trigonometric Substitution

: \int_{0}^{a r c s i n (x)} 1 d u

= \int_{0}^{a r c s i n (x)} 1 d u

Integral of a constant:

\int a d x = a x

= [1 \cdot u]_{0}^{a r c s i n (x)}

Simplify

= [u]_{0}^{a r c s i n (x)}

Compute the boundaries:

arcsin (x)

= arcsin (x)

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\int_{0}^{1} e^{x} cos (e^{x}) d x

\int_{6}^{8} \frac{1}{x ^{2}} d x

\int_{\frac{π}{4}}^{π} cos (x) d x

What is the integral from 0 to x of (1-t^2)^{-1/2} ?
The integral from 0 to x of (1-t^2)^{-1/2} is arcsin(x)