What is the integral from 0 to 1/3 of 1/(1+9x^2) ?

The integral from 0 to 1/3 of 1/(1+9x^2) is pi/(12)

Popular >

integral from 0 to 1/3 of 1/(1+9x^2)

\int_{0}^{\frac{1}{3}} \frac{1}{1 + 9 x ^{2}} d x

\frac{π}{12}

Decimal Notation

0.26179 \dots

Solution steps

\int_{0}^{\frac{1}{3}} \frac{1}{1 + 9 x ^{2}} d x

Apply Integral Substitution

= \int_{0}^{1} \frac{1}{3 ( 1 + u ^{2} )} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int_{0}^{1} \frac{1}{1 + u ^{2}} d u

Use the common integral:

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

= \frac{1}{3} [arctan (u)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= \frac{1}{3} \cdot \frac{π}{4}

Simplify

= \frac{π}{12}

\int_{- 1}^{1} \frac{e ^{x} - e ^{- x}}{2} d x

\int_{0}^{3} 1 - x^{2} d x

\int_{4}^{\infty} \frac{1}{x ^{2}} d x

\int_{0}^{4} (4 - x^{2}) - (4 - 4 x) d x

\int_{5}^{x^{2}} 2 1 + x^{2} d x

What is the integral from 0 to 1/3 of 1/(1+9x^2) ?
The integral from 0 to 1/3 of 1/(1+9x^2) is pi/(12)