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`int arcsec(2x) dx` Find the indefinite integral

Recall that indefinite integral follows `int f(x) dx = F(x) +C` where:


`f(x)` as the integrand function


`F(x)` as the antiderivative of `f(x)`


`C` as the constant of integration.


The given  integral problem: `int arcsec(2x)dx` resembles one of the formulas from the integration table. We follow the integral formula for inverse secant function as:


`int arcsec(u) du =u*arcsec(u) - ln|u+sqrt(u^2-1)|+C `


                       or    `u*arcsec(u)-cosh^(-1)|x|+C`


For easier comparison, we may apply u-substitution by letting: `u = 2x` then `du = 2 dx` or `(du)/2 =dx` .


Plug-in the values, we get:


`int arcsec(2x)dx=int arcsec(u) * (du)/2`


Apply the basic integration property: `int c*f(x) dx = c int f(x) dx` .


`int arcsec(u) * (du)/2= 1/2int arcsec(u) du`


Apply aforementioned integral formula for inverse secant function:


`1/2int arcsec(u) du =1/2*[u*arcsec(u) - ln|u+sqrt(u^2-1)|]+C`


                                   `=(u*arcsec(u))/2 -( ln|u+sqrt(u^2-1)|)/2+C`


Plug-in `u =2x` on `(u*arcsec(u))/2 -( ln|u+sqrt(u^2-1)|)/2+C` , we get the indefinite integral as:


`int arcsec(2x)dx =(2x*arcsec(2x))/2 -(ln|2x+sqrt((2x)^2-1)|)/2+C`


                                   `=xarcsec(2x) -(ln|2x+sqrt(4x^2-1)|)/2+C`



Another form of indefinite integral:


`1/2int arcsec(u) du= 1/2 *[u*arcsec(u)-cosh^(-1)|x|]+C`


                              `=(u*arcsec(u))/2-(cosh^(-1)|x|)/2+C`


Plug-in `u =2x` on  `(u*arcsec(u))/2-(cosh^(-1)|x|)/2+C`  , we get:


`int arcsec(2x)dx =(2x*arcsec(2x))/2-(cosh^(-1)|2x|)/2+C`


                                   ` =x*arcsec(2x)-(cosh^(-1)|2x|)/2+C`

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