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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

arctan (\frac{1}{2})

Decimal

0.46364 \dots

Solution steps

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

Use the common integral:

\int \frac{1}{1 + x ^{2}} d x = arctan (x)

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

Compute the boundaries:

arctan (\frac{1}{2})

= arctan (\frac{1}{2})

\int_{0}^{2 π} 4 d t

\int_{- 2}^{3} (4 x - 5) d x

\int_{0}^{3} \frac{1}{3} x^{3} d x

\int_{x}^{1} cos (t^{2}) d t

\int_{- 1}^{1} \frac{( x + 3 ) ^{3} e ^{x}}{2} d x

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