What is the integral of arcsec(mx) ?

The integral of arcsec(mx) is xarcsec(mx)-1/m ln|mx+sqrt(m^2x^2-1)|+C

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integral of arcsec(mx)

\int arcsec (m x) d x

x arcsec (m x) - \frac{1}{m} ln m x + m^{2} x^{2} - 1 + C

Solution steps

\int arcsec (m x) d x

Apply u-substitution

: \int \frac{arcsec ( u )}{m} d u

= \int \frac{arcsec ( u )}{m} d u

Take the constant out:

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

= \frac{1}{m} \cdot \int arcsec (u) d u

Apply Integration By Parts

: u arcsec (u) - \int \frac{u}{u ^{2} u ^{2} - 1} d u

= \frac{1}{m} (u arcsec (u) - \int \frac{u}{u ^{2} u ^{2} - 1} d u)

\int \frac{u}{u ^{2} u ^{2} - 1} d u = ln u + u^{2} - 1

= \frac{1}{m} (u arcsec (u) - ln u + u^{2} - 1)

Substitute back

u = m x

= \frac{1}{m} (m x arcsec (m x) - ln m x + (m x)^{2} - 1)

Simplify

\frac{1}{m} (m x arcsec (m x) - ln m x + (m x)^{2} - 1) : x arcsec (m x) - \frac{1}{m} ln m x + m^{2} x^{2} - 1

= x arcsec (m x) - \frac{1}{m} ln m x + m^{2} x^{2} - 1

Add a constant to the solution

= x arcsec (m x) - \frac{1}{m} ln m x + m^{2} x^{2} - 1 + C

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What is the integral of arcsec(mx) ?
The integral of arcsec(mx) is xarcsec(mx)-1/m ln|mx+sqrt(m^2x^2-1)|+C