What is the integral from-a to a of a/(pi(x^2+a^2)) ?

The integral from-a to a of a/(pi(x^2+a^2)) is 1/2

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integral from-a to a of a/(pi(x^2+a^2))

\int_{- a}^{a} \frac{a}{π ( x ^{2} + a ^{2} )} d x

\frac{1}{2}

Decimal

0.5

Solution steps

\int_{- a}^{a} \frac{a}{π ( x ^{2} + a ^{2} )} d x

Take the constant out:

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

= \frac{a}{π} \cdot \int_{- a}^{a} \frac{1}{x ^{2} + a ^{2}} d x

Apply Integral Substitution

= \frac{a}{π} \cdot \int_{- 1}^{1} \frac{1}{a ( u ^{2} + 1 )} d u

Take the constant out:

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

= \frac{a}{π} \cdot \frac{1}{a} \cdot \int_{- 1}^{1} \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= \frac{a}{π} \cdot \frac{1}{a} [arctan (u)]_{- 1}^{1}

Simplify

\frac{a}{π} \cdot \frac{1}{a} [arctan (u)]_{- 1}^{1} : \frac{1}{π} [arctan (u)]_{- 1}^{1}

= \frac{1}{π} [arctan (u)]_{- 1}^{1}

Compute the boundaries:

\frac{π}{2}

= \frac{1}{π} \cdot \frac{π}{2}

Simplify

= \frac{1}{2}

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What is the integral from-a to a of a/(pi(x^2+a^2)) ?
The integral from-a to a of a/(pi(x^2+a^2)) is 1/2