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

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

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

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

arcsin (\frac{b}{3})

Solution steps

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

Apply Trigonometric Substitution

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

= \int_{0}^{a r c s i n (\frac{b}{3})} 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}{3})}

Simplify

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

Compute the boundaries:

arcsin (\frac{b}{3})

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

\int - 9.8 x d x

(4 t + 2 y + 5) d t - d y = 0

\int \frac{sec ( x ) tan ( x )}{sec ( x ) - 3} d x

x^{2} - 4, [- 2, 0]

\frac{1}{2 - x}

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