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 Notation

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

= \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}

f (t) = ln (cos (t))

f (x) = e^{x} cos (x), (0, 1)

\frac{\partial}{\partial x} (ln (7 x^{2} + y^{2} + 6))

8 x^{3}

x y^{^{'}} + 2 y = - 2 e^{x}

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