What is the integral from 0 to b of 1/(sqrt(16-x^2)) ?

The integral from 0 to b of 1/(sqrt(16-x^2)) is arcsin(b/4)

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integral from 0 to b of 1/(sqrt(16-x^2))

\int_{0}^{b} \frac{1}{16 - x ^{2}} d x

arcsin (\frac{b}{4})

Solution steps

\int_{0}^{b} \frac{1}{16 - x ^{2}} d x

Apply Trigonometric Substitution

: \int_{0}^{a r c s i n (\frac{b}{4})} 1 d u

= \int_{0}^{a r c s i n (\frac{b}{4})} 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}{4})}

Simplify

= [u]_{0}^{a r c s i n (\frac{b}{4})}

Compute the boundaries:

arcsin (\frac{b}{4})

= arcsin (\frac{b}{4})

\frac{s}{s ^{2} + 2 s + 1}

\frac{d}{d x} (co tan (x) - tan (x))

\int \frac{x ^{2} - 2 x}{x} d x

\frac{d}{d t} (\frac{2 2}{3} t^{\frac{3}{2}})

What is the integral from 0 to b of 1/(sqrt(16-x^2)) ?
The integral from 0 to b of 1/(sqrt(16-x^2)) is arcsin(b/4)