Solving #intsec^3(2x)# #dx#
Apply u-substitution
Let #color(blue)(u=2x#
Thus, #color(red)(du=2dx->1/2du=dx#
Rewriting the integral...
#int color(red)(1/2)sec^3(color(blue)u)# #color(red)(du#
Take out the constant:# (1/2)#
#color(red)(1/2)intsec^3(color(blue)u)color(red)(du#
Solving #intsec^3u# #du# we must apply the secant reduction formula which states:
#intsec^n u# #du = (n-2)/(n-1)intsec^(n-2)u# #du+(sec^(n-2)utanu)/(n-1):n>=3#
Thus with #n=3#
#intsec^3u# #du = (3-2)/(3-1)intsec^(3-2)u# #du+(sec^(3-2)utanu)/(3-1)#
#intsec^3u# #du=1/2intsecu# #du+(secutanu)/(2)#
We now have the standard integral #intsecu# #du# which yields #ln|tanu+secu|#
Thus,
#intsec^3u# #dx=(ln|tanu+secu|)/2+(secutanu)/(2)+"C"#
#
Simplifying, we can factor out a #1/2# thus leading to the complete solution to the integral
#intsec^3u# #dx=1/2{(ln|tanu+secu|)+(secutanu)}+"C"#
Going back to our original integral:
#color(red)(1/2)intsec^3(color(blue)u)color(red)(du#
We can now write:
#1/2{1/2[(ln|tanu+secu|)+secutanu]}+"C"#
Simplify:
#((ln|tanu+secu|)+secutanu)/4+"C"#
And lastly substituting back #2x# for #u# we get:
#((ln|tan(2x)+sec(2x)|)+sec(2x)tan(2x))/4+"C"#
