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

The integral from 0 to tan(x) of ()/(1+t^2) is (tan(x)dt)/(t^2+1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

\int_{0}^{t a n (x)} \frac{d t}{1 + t ^{2}} d x

\frac{tan ( x ) d t}{t ^{2} + 1}

Solution steps

\int_{0}^{t a n (x)} \frac{d t}{1 + t ^{2}} d x

Integral of a constant:

\int a d x = a x

= [\frac{d t}{1 + t ^{2}} x]_{0}^{t a n (x)}

Compute the boundaries:

\frac{d t}{t ^{2} + 1} tan (x)

= \frac{d t}{t ^{2} + 1} tan (x)

Simplify

= \frac{tan ( x ) d t}{t ^{2} + 1}

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