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

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

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 of 1/(1+t^2)

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

\frac{π}{4}

Decimal

0.78539 \dots

Solution steps

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

Use the common integral:

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

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

Compute the boundaries:

\frac{π}{4}

= \frac{π}{4}

\int (3 + 2 cos (θ))^{2} d θ

x \to 4 lim (\frac{2}{x - 4})

\frac{\partial}{\partial x} (tan (6 + 2 x^{2} y^{4} z^{2}))

\frac{d}{d x} ((x^{3} + 1) (2 x^{2} + 8 x - 5))

\int \frac{1}{1 + 2 x ^{2}} d x

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