What is the integral from 0 to 2 of 1/(sqrt(16-x^2)) ?

The integral from 0 to 2 of 1/(sqrt(16-x^2)) is pi/6

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integral from 0 to 2 of 1/(sqrt(16-x^2))

\int_{0}^{2} \frac{1}{16 - x ^{2}} d x

\frac{π}{6}

Decimal Notation

0.52359 \dots

Solution steps

\int_{0}^{2} \frac{1}{16 - x ^{2}} d x

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{6}} 1 d u

= \int_{0}^{\frac{π}{6}} 1 d u

Integral of a constant:

\int a d x = a x

= [1 \cdot u]_{0}^{\frac{π}{6}}

Simplify

= [u]_{0}^{\frac{π}{6}}

Compute the boundaries:

\frac{π}{6}

= \frac{π}{6}

\frac{d}{d θ} (3 sin (2 θ))

n = 1 \sum \infty cos (4 n)

\frac{d}{d x} (a e^{a x})

\frac{d}{d x} ((x^{3} + 2 x - 7) (3 + x - x^{2}))

\frac{d}{d x} ((x + \frac{1}{x})^{- 1})

What is the integral from 0 to 2 of 1/(sqrt(16-x^2)) ?
The integral from 0 to 2 of 1/(sqrt(16-x^2)) is pi/6