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

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

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

\int_{0}^{b} \frac{1}{4 - x ^{2}} d x

arcsin (\frac{b}{2})

Solution steps

\int_{0}^{b} \frac{1}{4 - x ^{2}} d x

Apply Trigonometric Substitution

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

= \int_{0}^{a r c s i n (\frac{b}{2})} 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}{2})}

Simplify

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

Compute the boundaries:

arcsin (\frac{b}{2})

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

y = 7 cos (x), at x = \frac{π}{3}

y^{^{'}} = 0.4 (10 - x) (5 - x)

\int 5 cos^{2} (x) tan^{3} (x) d x

\frac{d}{d x} (3 arctan (x + 1 + x^{2}))

\int \frac{x}{x + 1 + 1} d x

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