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

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

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

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

\frac{π}{2}

Decimal

1.57079 \dots

Solution steps

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

Simplify

(1 - x^{2})^{\frac{- 1}{2}} : \frac{1}{( 1 - x ^{2} ) ^{\frac{1}{2}}}

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

Apply exponent rule:

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

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

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

Apply Trigonometric Substitution

= \int_{0}^{\frac{π}{2}} 1 d u

Integral of a constant:

\int a d x = a x

= [1 \cdot u]_{0}^{\frac{π}{2}}

Simplify

= [u]_{0}^{\frac{π}{2}}

Compute the boundaries:

\frac{π}{2}

= \frac{π}{2}

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