What is the integral of arcsec(2x) ?

The integral of arcsec(2x) is xarcsec(2x)-1/2 ln|2x+sqrt(4x^2-1)|+C

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integral of arcsec(2x)

\int arcsec (2 x) d x

x arcsec (2 x) - \frac{1}{2} ln 2 x + 4 x^{2} - 1 + C

Solution steps

\int arcsec (2 x) d x

Apply u-substitution

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

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

Take the constant out:

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

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

Substitute back

u = 2 x

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

Simplify

\frac{1}{2} (2 x arcsec (2 x) - ln 2 x + (2 x)^{2} - 1) : x arcsec (2 x) - \frac{1}{2} ln 2 x + 4 x^{2} - 1

= x arcsec (2 x) - \frac{1}{2} ln 2 x + 4 x^{2} - 1

Add a constant to the solution

= x arcsec (2 x) - \frac{1}{2} ln 2 x + 4 x^{2} - 1 + C

x \to - 4 lim (x + 3)

\int \frac{x}{x ( x - 2 ) ( x + 3 )} d x

\int cos (3 y) d y

x \to \infty lim ((cos (\frac{1}{x}))^{\frac{1}{x ^{2}}})

n = 3 \sum \infty \frac{n x ^{n}}{n + 2}

What is the integral of arcsec(2x) ?
The integral of arcsec(2x) is xarcsec(2x)-1/2 ln|2x+sqrt(4x^2-1)|+C