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

The integral from 2 to 4 of 1/(sqrt(16-t^2)) is pi/3

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

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

\frac{π}{3}

Decimal Notation

1.04719 \dots

Solution steps

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

Apply Trigonometric Substitution

: \int_{\frac{π}{6}}^{\frac{π}{2}} 1 d u

= \int_{\frac{π}{6}}^{\frac{π}{2}} 1 d u

Integral of a constant:

\int a d x = a x

= [1 \cdot u]_{\frac{π}{6}}^{\frac{π}{2}}

Simplify

= [u]_{\frac{π}{6}}^{\frac{π}{2}}

Compute the boundaries:

\frac{π}{3}

= \frac{π}{3}

x \to 2 lim (2 x - 2)

\int \frac{x ^{3}}{x ^{2} - 2 x + 4} d x

\int \frac{sec ( x )}{x} d x

y^{^{''}} - 2 y^{^{'}} - 3 y = - 3 t e^{- t}

x \to 1 lim (x - 2)

What is the integral from 2 to 4 of 1/(sqrt(16-t^2)) ?
The integral from 2 to 4 of 1/(sqrt(16-t^2)) is pi/3