de

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Beliebt Rechnen >

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

Lösung

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

Lösung

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

+1

Dezimale

0.14132 \dots

Schritte zur Lösung

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

Wende die partielle Integration an

= [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}}

Vereinfache

[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}}

Berechne die Grenzen:

\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))

Graph

Beliebte Beispiele

(\partial)/(\partial w)(warctan(uv))

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

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

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

sum from n=1 to infinity of 9

n = 1 \sum \infty 9

limit as x approaches 0-of |(8x)/x |

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

integral of sin((nx)/2)

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