What is the integral from 0 to 1/2 of 1/(36x^2+9) ?

The integral from 0 to 1/2 of 1/(36x^2+9) is pi/(72)

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integral from 0 to 1/2 of 1/(36x^2+9)

\int_{0}^{\frac{1}{2}} \frac{1}{36 x ^{2} + 9} d x

\frac{π}{72}

Solution steps

\int_{0}^{\frac{1}{2}} \frac{1}{36 x ^{2} + 9} d x

Apply Integral Substitution

= \int_{0}^{1} \frac{1}{2 ( 9 u ^{2} + 9 )} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{1} \frac{1}{9 u ^{2} + 9} d u

\frac{1}{9 u ^{2} + 9} = \frac{1}{9} \frac{1}{u ^{2} + 1}

= \frac{1}{2} \cdot \int_{0}^{1} \frac{1}{9} \cdot \frac{1}{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}{2} \cdot \frac{1}{9} \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}{2} \cdot \frac{1}{9} [arctan (u)]_{0}^{1}

Simplify

\frac{1}{2} \cdot \frac{1}{9} [arctan (u)]_{0}^{1} : \frac{1}{18} [arctan (u)]_{0}^{1}

= \frac{1}{18} [arctan (u)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= \frac{1}{18} \cdot \frac{π}{4}

Simplify

= \frac{π}{72}

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What is the integral from 0 to 1/2 of 1/(36x^2+9) ?
The integral from 0 to 1/2 of 1/(36x^2+9) is pi/(72)