What is the integral from 0 to 5/4 pi of 4/(25+x^2) ?

The integral from 0 to 5/4 pi of 4/(25+x^2) is 4/5 arctan(pi/4)

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integral from 0 to 5/4 pi of 4/(25+x^2)

\int_{0}^{\frac{5}{4} π} \frac{4}{25 + x ^{2}} d x

\frac{4}{5} arctan (\frac{π}{4})

Decimal

0.53261 \dots

Solution steps

\int_{0}^{\frac{5}{4} π} \frac{4}{25 + x ^{2}} d x

Take the constant out:

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

= 4 \cdot \int_{0}^{\frac{5}{4} π} \frac{1}{25 + x ^{2}} d x

Apply Integral Substitution

= 4 \cdot \int_{0}^{\frac{π}{4}} \frac{1}{5 ( u ^{2} + 1 )} d u

Take the constant out:

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

= 4 \cdot \frac{1}{5} \cdot \int_{0}^{\frac{π}{4}} \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= 4 \cdot \frac{1}{5} [arctan (u)]_{0}^{\frac{π}{4}}

Simplify

4 \cdot \frac{1}{5} [arctan (u)]_{0}^{\frac{π}{4}} : \frac{4}{5} [arctan (u)]_{0}^{\frac{π}{4}}

= \frac{4}{5} [arctan (u)]_{0}^{\frac{π}{4}}

Compute the boundaries:

arctan (\frac{π}{4})

= \frac{4}{5} arctan (\frac{π}{4})

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What is the integral from 0 to 5/4 pi of 4/(25+x^2) ?
The integral from 0 to 5/4 pi of 4/(25+x^2) is 4/5 arctan(pi/4)