What is the integral from 0 to 3 of 1/(sqrt(36-x^2)) ?

The integral from 0 to 3 of 1/(sqrt(36-x^2)) is pi/6

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integral from 0 to 3 of 1/(sqrt(36-x^2))

\int_{0}^{3} \frac{1}{36 - x ^{2}} d x

\frac{π}{6}

Decimal Notation

0.52359 \dots

Solution steps

\int_{0}^{3} \frac{1}{36 - x ^{2}} d x

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{6}} 1 d u

= \int_{0}^{\frac{π}{6}} 1 d u

Integral of a constant:

\int a d x = a x

= [1 \cdot u]_{0}^{\frac{π}{6}}

Simplify

= [u]_{0}^{\frac{π}{6}}

Compute the boundaries:

\frac{π}{6}

= \frac{π}{6}

f (x) = (6 x + 2)^{3}

\int - \frac{x}{x - 1} d x

y = 13 + 2 x - x^{2}, y = 3 x + 1

\int \frac{x ^{4}}{1 + x ^{2}} d x

\frac{d}{d x} (\frac{x ^{4}}{4} + \frac{8}{9} x^{3} - x^{2} + 3 x - 8)

What is the integral from 0 to 3 of 1/(sqrt(36-x^2)) ?
The integral from 0 to 3 of 1/(sqrt(36-x^2)) is pi/6