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

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

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

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

\frac{π}{2}

Decimal

1.57079 \dots

Solution steps

\int_{0}^{1} \frac{2}{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

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{4}} 1 d u

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

Integral of a constant:

\int a d x = a x

= 2 [1 \cdot u]_{0}^{\frac{π}{4}}

Simplify

= 2 [u]_{0}^{\frac{π}{4}}

Compute the boundaries:

\frac{π}{4}

= 2 \cdot \frac{π}{4}

Simplify

= \frac{π}{2}

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

(x^{2} - 9 x + 9) 25 - x^{2}

x \to 0 lim (\frac{( 3 ^{x} - 2 ^{x} )}{x})

\int (e^{2 x} + e^{- 3 x}) d x

(ln (x))^{5}

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