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

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

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

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

\frac{π}{80}

Decimal

0.03926 \dots

Solution steps

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

y = sin^{2} (x) \frac{d y}{d x}

\int (x^{3} + x^{- 4} + x^{\frac{3}{4}}) d x

\int \frac{( sin ( x ) + cos ( x ) ) ^{2}}{sin ( x )} d x

\frac{1}{z ^{2} + 1}

x^{- 5} (1 + x^{- 1}), at x = 1

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