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

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

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

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

\frac{π}{2}

Decimal Notation

1.57079 \dots

Solution steps

\int_{0}^{9} \frac{1}{81 - 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{\partial}{\partial x} (6 x y - 5 x^{2} y)

x, 3

\frac{d y}{d x}, x^{2} = \frac{( 3 x + 5 y )}{3 x - 5 y}

x \to - 2 + lim (\frac{x + 1}{x ^{2} + 2})

\frac{5}{( s + 1 ) ^{2}}

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