How do you integrate $\int {sec}^{2} (2 x - 1)$ $int sec^2(2x-1)$ ?

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How do you integrate $\int {sec}^{2} (2 x - 1)$ $int sec^2(2x-1)$ ?

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Answer 1 · 2017-01-10T22:14:43

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

Explanation:

Note that $\frac{d}{d x} tan (x) = {sec}^{2} (x)$ $d/dxtan(x)=sec^2(x)$ . This implies that $\int {sec}^{2} (x) d x = tan (x) + C$ $intsec^2(x)dx=tan(x)+C$ .

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

$\int {sec}^{2} (2 x - 1) d x = \frac{1}{2} \int {sec}^{2} (2 x - 1) (2 d x) = \frac{1}{2} \int {sec}^{2} (u) d u$ $intsec^2(2x-1)dx=1/2intsec^2(2x-1)(2dx)=1/2intsec^2(u)du$

As we saw before, $\int {sec}^{2} (u) d u = tan (u) + C$ $intsec^2(u)du=tan(u)+C$ :

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

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

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

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