How do you integrate #int sec^2(2x-3)# using substitution?

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Answer 1 · 2018-04-04T02:54:10

#intsec^2(2x-3)dx=1/2tan(2x-3)+"C"#

Let #u=2x-3#

Then, #du=2dx=>color(red)(1/2du=dx#

So we can rewrite the given integral as:

#color(red)(1/2)intsec^2ucolor(red)(du#

Solving the integral we get...

#1/2tanu#

Reversing the substitution...

#1/2tan(2x-3)+"C"#