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

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

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

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

\frac{π}{2}

Decimal Notation

1.57079 \dots

Solution steps

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

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}

f (x) = - \frac{1}{x ^{2}}

y = - x^{2} + 1, (0, 1)

\frac{7 x + 1}{2 x - 3}

\int_{0}^{1} x^{3} ln (x^{8}) d x

f (x) = \frac{x}{x + 3}, at x = 1

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