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

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

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

\int_{0}^{6} \frac{1}{36 - x ^{2}} d x

\frac{π}{2}

Decimal Notation

1.57079 \dots

Solution steps

\int_{0}^{6} \frac{1}{36 - 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}

\int (y + 2) d y

x \to 3 lim ((x) + (4 - x))

f (x) = \frac{1 - x}{x}

\frac{d}{d x} (x sin (x) cos (x))

\int x^{10} d x

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