What is the integral from 0 to 9 of 1/(sqrt(81-t^2)) ?

The integral from 0 to 9 of 1/(sqrt(81-t^2)) is pi/2

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integral from 0 to 9 of 1/(sqrt(81-t^2))

\int_{0}^{9} \frac{1}{81 - t ^{2}} d t

\frac{π}{2}

Solution steps

\int_{0}^{9} \frac{1}{81 - 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}

\int sin (φ) d φ

\frac{d}{d x} (x + 1 (x - 1))

x \to 0 lim (\frac{1}{x ^{13}})

(x^{2} + 1) \frac{d y}{d x} = x^{2} (y - 1)

\frac{d}{d x} (x y 1 - \frac{x ^{2}}{4} - \frac{y ^{2}}{9})

What is the integral from 0 to 9 of 1/(sqrt(81-t^2)) ?
The integral from 0 to 9 of 1/(sqrt(81-t^2)) is pi/2