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 Notation

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

= \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

\frac{\partial}{\partial x} (2 x + 1 + y e^{- x})

\int cos (x) (8 + 7 sin^{2} (x)) d x

\int e^{- 20 x} d x

y = 10 x sin (x)

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