What is the integral from 0 to 3 of 1/(sqrt(9-x^2)) ?

The integral from 0 to 3 of 1/(sqrt(9-x^2)) is pi/2

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integral from 0 to 3 of 1/(sqrt(9-x^2))

\int_{0}^{3} \frac{1}{9 - x ^{2}} d x

\frac{π}{2}

Decimal Notation

1.57079 \dots

Solution steps

\int_{0}^{3} \frac{1}{9 - 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} (- A B^{- x} + C)

\frac{d}{d x} (2 x (1 - x)^{3} - 3 x^{2} (1 - x)^{2})

\int \frac{x + 1}{x + 2} d x

\int \frac{x}{x + 20} d x

(x + x) (\frac{d y}{d x}) = y + y

What is the integral from 0 to 3 of 1/(sqrt(9-x^2)) ?
The integral from 0 to 3 of 1/(sqrt(9-x^2)) is pi/2