What is the integral from 4 to 8 of 1/(sqrt(64-x^2)) ?

The integral from 4 to 8 of 1/(sqrt(64-x^2)) is pi/3

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integral from 4 to 8 of 1/(sqrt(64-x^2))

\int_{4}^{8} \frac{1}{64 - x ^{2}} d x

\frac{π}{3}

Decimal

1.04719 \dots

Solution steps

\int_{4}^{8} \frac{1}{64 - x ^{2}} d x

Apply Trigonometric Substitution

: \int_{\frac{π}{6}}^{\frac{π}{2}} 1 d u

= \int_{\frac{π}{6}}^{\frac{π}{2}} 1 d u

Integral of a constant:

\int a d x = a x

= [1 \cdot u]_{\frac{π}{6}}^{\frac{π}{2}}

Simplify

= [u]_{\frac{π}{6}}^{\frac{π}{2}}

Compute the boundaries:

\frac{π}{3}

= \frac{π}{3}

\int_{0}^{0.4} x^{2} (1 - x)^{2} d x

\int_{- 1}^{3} 2 x + 8 d x

\int_{1}^{\infty} x sin (x) d x

\int_{0}^{\frac{1}{2}} 3 x - 6 x^{2} d x

\int_{1}^{64} x^{\frac{- 5}{6}} d x

What is the integral from 4 to 8 of 1/(sqrt(64-x^2)) ?
The integral from 4 to 8 of 1/(sqrt(64-x^2)) is pi/3