What is the integral from 0 to pi/8 of arctan(2x) ?

The integral from 0 to pi/8 of arctan(2x) is 1/4 (pi/2 arctan(pi/4)-ln((pi^2)/(16)+1))

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integral from 0 to pi/8 of arctan(2x)

\int_{0}^{\frac{π}{8}} arctan (2 x) d x

\frac{1}{4} (\frac{π}{2} arctan (\frac{π}{4}) - ln (\frac{π ^{2}}{16} + 1))

Decimal Notation

0.14132 \dots

Solution steps

\int_{0}^{\frac{π}{8}} arctan (2 x) d x

Apply Integration By Parts

: [x arctan (2 x) - \int \frac{2 x}{4 x ^{2} + 1} d x]_{0}^{\frac{π}{8}}

= [x arctan (2 x) - \int \frac{2 x}{4 x ^{2} + 1} d x]_{0}^{\frac{π}{8}}

\int \frac{2 x}{4 x ^{2} + 1} d x = \frac{1}{4} ln 4 x^{2} + 1

= [x arctan (2 x) - \frac{1}{4} ln 4 x^{2} + 1]_{0}^{\frac{π}{8}}

Simplify

[x arctan (2 x) - \frac{1}{4} ln 4 x^{2} + 1]_{0}^{\frac{π}{8}} : [\frac{1}{4} (4 x arctan (2 x) - ln 4 x^{2} + 1)]_{0}^{\frac{π}{8}}

= [\frac{1}{4} (4 x arctan (2 x) - ln 4 x^{2} + 1)]_{0}^{\frac{π}{8}}

Compute the boundaries:

\frac{1}{4} (\frac{π}{2} arctan (\frac{π}{4}) - ln (\frac{π ^{2}}{16} + 1))

= \frac{1}{4} (\frac{π}{2} arctan (\frac{π}{4}) - ln (\frac{π ^{2}}{16} + 1))

\frac{\partial}{\partial w} (w arctan (uv))

y^{^{''}} + 2 y - 1 (y^{^{'}}) 2 = y^{^{'}}

n = 1 \sum \infty 9

x \to 0 - lim (\frac{8 x}{x})

\int sin (\frac{n x}{2}) d x

What is the integral from 0 to pi/8 of arctan(2x) ?
The integral from 0 to pi/8 of arctan(2x) is 1/4 (pi/2 arctan(pi/4)-ln((pi^2)/(16)+1))