What is the integral from 0 to 2 of 2/(sqrt(8-x^2)) ?

The integral from 0 to 2 of 2/(sqrt(8-x^2)) is pi/2

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integral from 0 to 2 of 2/(sqrt(8-x^2))

\int_{0}^{2} \frac{2}{8 - x ^{2}} d x

\frac{π}{2}

Decimal Notation

1.57079 \dots

Solution steps

\int_{0}^{2} \frac{2}{8 - 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}^{2} \frac{1}{8 - 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}

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What is the integral from 0 to 2 of 2/(sqrt(8-x^2)) ?
The integral from 0 to 2 of 2/(sqrt(8-x^2)) is pi/2