What is the integral from 0 to 2 of 1/(4+t^2) ?

The integral from 0 to 2 of 1/(4+t^2) is pi/8

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integral from 0 to 2 of 1/(4+t^2)

\int_{0}^{2} \frac{1}{4 + t ^{2}} d t

\frac{π}{8}

Decimal

0.39269 \dots

Solution steps

\int_{0}^{2} \frac{1}{4 + t ^{2}} d t

Apply Integral Substitution

= \int_{0}^{1} \frac{1}{2 ( u ^{2} + 1 )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{0}^{1} \frac{1}{u ^{2} + 1} d u

Use the common integral:

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= \frac{1}{2} [arctan (u)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= \frac{1}{2} \cdot \frac{π}{4}

Simplify

= \frac{π}{8}

\int_{4}^{8} 1

\int_{0}^{1} \frac{x}{8} d x

\int_{4}^{2} 1

\int_{- \infty}^{0} 20 z e^{2 z} d z

\int_{1}^{3} e^{l n (x)} d x

What is the integral from 0 to 2 of 1/(4+t^2) ?
The integral from 0 to 2 of 1/(4+t^2) is pi/8