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 $u=2x-1$ :

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

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