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

The integral from 0 to 1 of 1/(sqrt(4-t^2)) is pi/6

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

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

\frac{π}{6}

Decimal

0.52359 \dots

Solution steps

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

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{6}} 1 d u

= \int_{0}^{\frac{π}{6}} 1 d u

Integral of a constant:

\int a d x = a x

= [1 \cdot u]_{0}^{\frac{π}{6}}

Simplify

= [u]_{0}^{\frac{π}{6}}

Compute the boundaries:

\frac{π}{6}

= \frac{π}{6}

y = \frac{5}{x ^{3}}

\frac{d}{d x} ((3 x)^{4 x})

\frac{d}{d x} (c x^{- 7})

\frac{d y}{d x} = \frac{sec ( y )}{( x + 2 ) ^{2}}

\int \frac{k}{x ^{4}} d x

What is the integral from 0 to 1 of 1/(sqrt(4-t^2)) ?
The integral from 0 to 1 of 1/(sqrt(4-t^2)) is pi/6