What is the integral from 0 to 4/9 of 1/(81x^2+16) ?

The integral from 0 to 4/9 of 1/(81x^2+16) is pi/(144)

Popular >

integral from 0 to 4/9 of 1/(81x^2+16)

\int_{0}^{\frac{4}{9}} \frac{1}{81 x ^{2} + 16} d x

\frac{π}{144}

Decimal Notation

0.02181 \dots

Solution steps

\int_{0}^{\frac{4}{9}} \frac{1}{81 x ^{2} + 16} d x

Apply Integral Substitution

= \int_{0}^{1} \frac{1}{36 ( 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}{36} \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}{36} [arctan (u)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= \frac{1}{36} \cdot \frac{π}{4}

Simplify

= \frac{π}{144}

\int_{0}^{l} x sin (\frac{nπ x}{l}) d x

\int_{0}^{2 x} sin (x) d x

\int_{0}^{π} (2 x - 1) cos (3 x) d x

\int_{- π}^{π} 2 sin (x) d x

\int_{0}^{8} x^{3} d x

What is the integral from 0 to 4/9 of 1/(81x^2+16) ?
The integral from 0 to 4/9 of 1/(81x^2+16) is pi/(144)