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

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

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

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

π

Decimal Notation

3.14159 \dots

Solution steps

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

Apply Trigonometric Substitution

: \int_{- \frac{π}{2}}^{\frac{π}{2}} 1 d u

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

Integral of a constant:

\int a d x = a x

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

Simplify

= [u]_{- \frac{π}{2}}^{\frac{π}{2}}

Compute the boundaries:

π

= π

\int \frac{1}{( 3 x + 1 )} d x

x \to 3 + lim (\frac{x + 4}{x - 3})

f (x) = cot^{2} (cos (θ))

\int \frac{( 2 x - 5 ) ^{2}}{3} d x

\int_{0}^{\frac{π}{6}} 3 cos^{5} (3 x) d x

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