What is the integral from 0 to 8 of 1/(sqrt(64-x^2)) ?

The integral from 0 to 8 of 1/(sqrt(64-x^2)) is pi/2

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integral from 0 to 8 of 1/(sqrt(64-x^2))

\int_{0}^{8} \frac{1}{64 - x ^{2}} d x

\frac{π}{2}

Decimal Notation

1.57079 \dots

Solution steps

\int_{0}^{8} \frac{1}{64 - x ^{2}} d x

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{2}} 1 d u

= \int_{0}^{\frac{π}{2}} 1 d u

Integral of a constant:

\int a d x = a x

= [1 \cdot u]_{0}^{\frac{π}{2}}

Simplify

= [u]_{0}^{\frac{π}{2}}

Compute the boundaries:

\frac{π}{2}

= \frac{π}{2}

\frac{d}{d x} (e^{3 x} sin (3 x))

x \to \infty lim (x^{2} - 2)

\int y^{\frac{2}{3}} d y

\frac{d}{d x} (x^{l n (3 x)})

\int \frac{25}{27} sin^{2} (x) d x

What is the integral from 0 to 8 of 1/(sqrt(64-x^2)) ?
The integral from 0 to 8 of 1/(sqrt(64-x^2)) is pi/2