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

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Answer 1 · 2018-04-04T02:54:10

$\int {sec}^{2} (2 x - 3) d x = \frac{1}{2} tan (2 x - 3) + C$ $intsec^2(2x-3)dx=1/2tan(2x-3)+"C"$

Let $u = 2 x - 3$ $u=2x-3$

Then, $d u = 2 d x \Rightarrow \frac{1}{2} d u = d x$ $du=2dx=>color(red)(1/2du=dx$

So we can rewrite the given integral as:

$\frac{1}{2} \int {sec}^{2} u d u$ $color(red)(1/2)intsec^2ucolor(red)(du$

Solving the integral we get...

$\frac{1}{2} tan u$ $1/2tanu$

Reversing the substitution...

$\frac{1}{2} tan (2 x - 3) + C$ $1/2tan(2x-3)+"C"$

How do you integrate ∫sec2(2x−3)int sec^2(2x-3) using substitution?