What is the integral from 0 to 2 of 1/(sqrt(25-x^2)) ?

The integral from 0 to 2 of 1/(sqrt(25-x^2)) is arcsin(2/5)

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integral from 0 to 2 of 1/(sqrt(25-x^2))

\int_{0}^{2} \frac{1}{25 - x ^{2}} d x

arcsin (\frac{2}{5})

Decimal Notation

0.41151 \dots

Solution steps

\int_{0}^{2} \frac{1}{25 - x ^{2}} d x

Apply Trigonometric Substitution

: \int_{0}^{a r c s i n (\frac{2}{5})} 1 d u

= \int_{0}^{a r c s i n (\frac{2}{5})} 1 d u

Integral of a constant:

\int a d x = a x

= [1 \cdot u]_{0}^{a r c s i n (\frac{2}{5})}

Simplify

= [u]_{0}^{a r c s i n (\frac{2}{5})}

Compute the boundaries:

arcsin (\frac{2}{5})

= arcsin (\frac{2}{5})

\int_{\frac{π}{5}}^{π} ∣ cos (x) ∣ d x

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

\int_{2}^{6} (8 t^{3} - 6 t^{- 2}) d t

\int_{- 1}^{0} (x^{5}) x^{3} + 1 d x

\int_{0}^{1} \frac{x ^{3}}{2} d x

What is the integral from 0 to 2 of 1/(sqrt(25-x^2)) ?
The integral from 0 to 2 of 1/(sqrt(25-x^2)) is arcsin(2/5)