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

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

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

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

\frac{π}{3}

Decimal Notation

1.04719 \dots

Solution steps

\int_{1}^{2} \frac{1}{4 - 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}

- 3 x^{2} + 2 x y + 2 y^{3} = 125, (3, 4)

\frac{d}{d x} y = ln (10 x^{2} - 8)

x^{3} - 3 x^{2} + 2 x - 2

\frac{d}{d x} (I n x)

f (x) = x^{3} - 2 x^{2} + 1, at x = 1

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