What is the integral from 0 to 4/5 of 1/(25x^2+16) ?

The integral from 0 to 4/5 of 1/(25x^2+16) is pi/(80)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 0 to 4/5 of 1/(25x^2+16)

\int_{0}^{\frac{4}{5}} \frac{1}{25 x ^{2} + 16} d x

\frac{π}{80}

Decimal

0.03926 \dots

Solution steps

\int_{0}^{\frac{4}{5}} \frac{1}{25 x ^{2} + 16} d x

Apply Integral Substitution

= \int_{0}^{1} \frac{1}{20 ( 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}{20} \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}{20} [arctan (u)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= \frac{1}{20} \cdot \frac{π}{4}

Simplify

= \frac{π}{80}

What is the integral from 0 to 4/5 of 1/(25x^2+16) ?
The integral from 0 to 4/5 of 1/(25x^2+16) is pi/(80)