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

The integral from 0 to 1 of 1/(sqrt(2-x^2)) is pi/4

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

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

\frac{π}{4}

Decimal Notation

0.78539 \dots

Solution steps

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

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{4}} 1 d u

= \int_{0}^{\frac{π}{4}} 1 d u

Integral of a constant:

\int a d x = a x

= [1 \cdot u]_{0}^{\frac{π}{4}}

Simplify

= [u]_{0}^{\frac{π}{4}}

Compute the boundaries:

\frac{π}{4}

= \frac{π}{4}

\int \frac{1}{x ^{2} + 4 x + 4} d x

\frac{\partial}{\partial y} (\frac{5 y}{x})

\frac{d}{d x} (\frac{4 x + 4}{3 x ^{\frac{2}{3}}})

\frac{\partial}{\partial y} (2 x + y^{3})

\frac{d}{d x} ((4 x^{3} + 1)^{3})

What is the integral from 0 to 1 of 1/(sqrt(2-x^2)) ?
The integral from 0 to 1 of 1/(sqrt(2-x^2)) is pi/4