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...

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

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