What is the integral from 0 to 1/2 of 1/(64x^2+16) ?

The integral from 0 to 1/2 of 1/(64x^2+16) is pi/(128)

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integral from 0 to 1/2 of 1/(64x^2+16)

\int_{0}^{\frac{1}{2}} \frac{1}{64 x ^{2} + 16} d x

\frac{π}{128}

Decimal

0.02454 \dots

Solution steps

\int_{0}^{\frac{1}{2}} \frac{1}{64 x ^{2} + 16} d x

Apply Integral Substitution

= \int_{0}^{1} \frac{1}{2 ( 16 u ^{2} + 16 )} 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}{16 u ^{2} + 16} d u

\frac{1}{16 u ^{2} + 16} = \frac{1}{16} \frac{1}{u ^{2} + 1}

= \frac{1}{2} \cdot \int_{0}^{1} \frac{1}{16} \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}{16} \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}{16} [arctan (u)]_{0}^{1}

Simplify

\frac{1}{2} \cdot \frac{1}{16} [arctan (u)]_{0}^{1} : \frac{1}{32} [arctan (u)]_{0}^{1}

= \frac{1}{32} [arctan (u)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= \frac{1}{32} \cdot \frac{π}{4}

Simplify

= \frac{π}{128}

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What is the integral from 0 to 1/2 of 1/(64x^2+16) ?
The integral from 0 to 1/2 of 1/(64x^2+16) is pi/(128)