What is the integral from 1 to x^2 of sec(t) ?

The integral from 1 to x^2 of sec(t) is ln|tan(x^2)+sec(x^2)|-ln(tan(1)+sec(1))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 1 to x^2 of sec(t)

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

ln tan (x^{2}) + sec (x^{2}) - ln (tan (1) + sec (1))

Solution steps

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

Use the common integral:

\int sec (t) d t = ln ∣ tan (t) + sec (t) ∣

= [ln ∣ tan (t) + sec (t) ∣]_{1}^{x^{2}}

Compute the boundaries:

ln tan (x^{2}) + sec (x^{2}) - ln (tan (1) + sec (1))

= ln tan (x^{2}) + sec (x^{2}) - ln (tan (1) + sec (1))

What is the integral from 1 to x^2 of sec(t) ?
The integral from 1 to x^2 of sec(t) is ln|tan(x^2)+sec(x^2)|-ln(tan(1)+sec(1))