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

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

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

\int_{3}^{6} \frac{1}{36 - x ^{2}} d x

\frac{π}{3}

Decimal

1.04719 \dots

Solution steps

\int_{3}^{6} \frac{1}{36 - x ^{2}} d x

Apply Trigonometric Substitution

= \int_{\frac{π}{6}}^{\frac{π}{2}} 1 d u

Integral of a constant:

\int a d x = a x

= [1 \cdot u]_{\frac{π}{6}}^{\frac{π}{2}}

Simplify

= [u]_{\frac{π}{6}}^{\frac{π}{2}}

Compute the boundaries:

\frac{π}{3}

= \frac{π}{3}

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