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

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

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

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

\frac{π}{2 2}

Decimal Notation

1.11072 \dots

Solution steps

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

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

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Simplify

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

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 21 \cdot \int_{0}^{1} \frac{1}{2 - x ^{2}} d x

Simplify

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

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{4}} 1 d u

= 2 \cdot 1 \cdot \int_{0}^{\frac{π}{4}} 1 d u

Integral of a constant:

\int a d x = a x

= 2 \cdot 1 \cdot [1 \cdot u]_{0}^{\frac{π}{4}}

Simplify

= 2 [u]_{0}^{\frac{π}{4}}

Compute the boundaries:

\frac{π}{4}

= 2 \frac{π}{4}

Simplify

= \frac{π}{2 2}

y = 3 x - 1

f (x) = e^{x} (x + x x)

\frac{d y}{d x} = y - y^{2}

f (x) = - x^{3}, (1, - 1)

x^{2} - 5 x + 5

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