What is the integral from 0 to 1 of 2/(sqrt(9-2x^2)) ?

The integral from 0 to 1 of 2/(sqrt(9-2x^2)) is sqrt(2)arcsin((sqrt(2))/3)

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integral from 0 to 1 of 2/(sqrt(9-2x^2))

\int_{0}^{1} \frac{2}{9 - 2 x ^{2}} d x

2 arcsin (\frac{2}{3})

Decimal Notation

0.69421 \dots

Solution steps

\int_{0}^{1} \frac{2}{9 - 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}{9 - 2 x ^{2}} d x

Apply Trigonometric Substitution

: \int_{0}^{a r c s i n (\frac{2}{3})} \frac{1}{2} d u

= 2 \cdot \int_{0}^{a r c s i n (\frac{2}{3})} \frac{1}{2} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int_{0}^{a r c s i n (\frac{2}{3})} 1 d u

Integral of a constant:

\int a d x = a x

= 2 \cdot \frac{1}{2} [1 \cdot u]_{0}^{a r c s i n (\frac{2}{3})}

Simplify

2 \cdot \frac{1}{2} [1 \cdot u]_{0}^{a r c s i n (\frac{2}{3})} : 2 [u]_{0}^{a r c s i n (\frac{2}{3})}

= 2 [u]_{0}^{a r c s i n (\frac{2}{3})}

Compute the boundaries:

arcsin (\frac{2}{3})

= 2 arcsin (\frac{2}{3})

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What is the integral from 0 to 1 of 2/(sqrt(9-2x^2)) ?
The integral from 0 to 1 of 2/(sqrt(9-2x^2)) is sqrt(2)arcsin((sqrt(2))/3)