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

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

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

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

arcsin (\frac{b}{5})

Solution steps

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

Apply Trigonometric Substitution

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

= \int_{0}^{a r c s i n (\frac{b}{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{b}{5})}

Simplify

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

Compute the boundaries:

arcsin (\frac{b}{5})

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

x \to - 4 lim (\frac{x ^{2} - 3}{4 - x})

2 x^{3} - x^{2} + 1

y^{^{''}} - 16 y^{^{'}} + 289 y = 4 t - 21 cos (17 t)

n = 1 \sum \infty n 2^{n}

\int \frac{1}{e ^{- x}} d x

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