What is the integral of xsec(4x^2-5) ?

The integral of xsec(4x^2-5) is 1/8 ln|tan(4x^2-5)+sec(4x^2-5)|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of xsec(4x^2-5)

\int x sec (4 x^{2} - 5) d x

\frac{1}{8} ln tan (4 x^{2} - 5) + sec (4 x^{2} - 5) + C

Solution steps

\int x sec (4 x^{2} - 5) d x

Apply u-substitution

= \int \frac{sec ( u )}{8} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{8} \cdot \int sec (u) d u

Use the common integral:

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

= \frac{1}{8} ln ∣ tan (u) + sec (u) ∣

Substitute back

u = 4 x^{2} - 5

= \frac{1}{8} ln tan (4 x^{2} - 5) + sec (4 x^{2} - 5)

Add a constant to the solution

= \frac{1}{8} ln tan (4 x^{2} - 5) + sec (4 x^{2} - 5) + C

What is the integral of xsec(4x^2-5) ?
The integral of xsec(4x^2-5) is 1/8 ln|tan(4x^2-5)+sec(4x^2-5)|+C