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

The integral from 0 to 1/2 of 1/(1+(2x)^2) is pi/8

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

\int_{0}^{\frac{1}{2}} \frac{1}{1 + ( 2 x ) ^{2}} d x

\frac{π}{8}

Decimal

0.39269 \dots

Solution steps

\int_{0}^{\frac{1}{2}} \frac{1}{1 + ( 2 x ) ^{2}} d x

Simplify

\frac{1}{1 + ( 2 x ) ^{2}} : \frac{1}{1 + 4 x ^{2}}

= \int_{0}^{\frac{1}{2}} \frac{1}{1 + 4 x ^{2}} d x

Apply Integral Substitution

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

Compute the boundaries:

\frac{π}{4}

= \frac{1}{2} \cdot \frac{π}{4}

Simplify

= \frac{π}{8}

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What is the integral from 0 to 1/2 of 1/(1+(2x)^2) ?
The integral from 0 to 1/2 of 1/(1+(2x)^2) is pi/8