What is the integral from 0 to 4 of 1/(sqrt(64-x^2)) ?

The integral from 0 to 4 of 1/(sqrt(64-x^2)) is pi/6

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integral from 0 to 4 of 1/(sqrt(64-x^2))

\int_{0}^{4} \frac{1}{64 - x ^{2}} d x

\frac{π}{6}

Solution steps

\int_{0}^{4} \frac{1}{64 - x ^{2}} d x

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}

\frac{\partial}{\partial x} (5 x y e^{- x^{2} - y^{2}})

e^{1}

k \to \infty lim \frac{( ( k + 1 ) ^{75} ( k + 4 ) )}{( k + 5 ) k ^{75}}

x \to \infty lim (ln (3))

x^{\frac{1}{3}}

What is the integral from 0 to 4 of 1/(sqrt(64-x^2)) ?
The integral from 0 to 4 of 1/(sqrt(64-x^2)) is pi/6