How do you integrate #int sec^2(2x-1)#?

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Answer 1 · 2017-01-10T22:14:43

#intsec^2(2x-1)dx=1/2tan(2x-1)+C#

Note that #d/dxtan(x)=sec^2(x)#. This implies that #intsec^2(x)dx=tan(x)+C#.

Apply the substitution #u=2x-1#, which implies that #du=2dx#:

#intsec^2(2x-1)dx=1/2intsec^2(2x-1)(2dx)=1/2intsec^2(u)du#

As we saw before, #intsec^2(u)du=tan(u)+C#:

#intsec^2(2x-1)dx=1/2tan(u)+C#

Returning to the original variable #x#, use #u=2x-1#:

#intsec^2(2x-1)dx=1/2tan(2x-1)+C#