What is the integral from-4 to 4 of 1/(sqrt(16-x^2)) ?

The integral from-4 to 4 of 1/(sqrt(16-x^2)) is pi

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integral from-4 to 4 of 1/(sqrt(16-x^2))

\int_{- 4}^{4} \frac{1}{16 - x ^{2}} d x

π

Decimal

3.14159 \dots

Solution steps

\int_{- 4}^{4} \frac{1}{16 - x ^{2}} d x

Apply Trigonometric Substitution

= \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

What is the integral from-4 to 4 of 1/(sqrt(16-x^2)) ?
The integral from-4 to 4 of 1/(sqrt(16-x^2)) is pi