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

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

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

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

\frac{π}{4}

Decimal Notation

0.78539 \dots

Solution steps

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

Take the constant out:

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

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

Apply Integral Substitution

= 2 \cdot \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

= 2 \cdot \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)

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

Simplify

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

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

Compute the boundaries:

\frac{π}{4}

= \frac{π}{4}

y^{^{'}} = \frac{2 x ( y - 1 )}{x ^{2} + 3}

\int 19 sin^{3} (x) cos^{2} (x) d x

f (x) = \frac{6}{x ^{2}}, at x = 1

y^{^{''}} - 3 y^{^{'}} - 10 y = 0

\frac{1}{( 1 + x ) ^{2}}, 1

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