How do you integrate $\int {sec}^{2} x tan x$ $int sec^2xtanx$ ?

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How do you integrate $\int {sec}^{2} x tan x$ $int sec^2xtanx$ ?

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Answer 1 · 2016-11-24T17:23:06

You have to remeber that ${sec}^{2} (x)$ $sec^2(x)$ is the derivative of $tan (x)$ $tan(x)$

Explanation:

So:

$\int {sec}^{2} x tan x d x = \int tan x d (tan x) = \frac{{tan}^{2} x}{2}$ $int sec^2 x tan x dx = int tan x d (tan x) = (tan^2 x ) / 2$

Answer 2 · 2016-11-24T17:24:49

$\frac{{sec}^{2} x}{2} + C$ $sec^2x/2+C$ or $\frac{{tan}^{2} x}{2} + C$ $tan^2x/2+C$

Explanation:

You can also see this this way:

$\int {sec}^{2} x tan x d x = \int sec x (sec x tan x) d x$ $intsec^2xtanxdx=intsecx(secxtanx)dx$

If $u = sec x$ $u=secx$ then $d u = sec x tan x d x$ $du=secxtanxdx$ so

$= \int u d u = \frac{u^{2}}{2} = \frac{{sec}^{2} x}{2} + C$ $=intudu=u^2/2=sec^2x/2+C$

This is equivalent to the other answer of $\frac{{tan}^{2} x}{2} + C$ $tan^2x/2+C$ because ${tan}^{2} x$ $tan^2x$ and ${sec}^{2} x$ $sec^2x$ are only a constant away from one another through the equality ${tan}^{2} x + 1 = {sec}^{2} x$ $tan^2x+1=sec^2x$ .

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How do you integrate $\int {sec}^{2} x tan x$ $int sec^2xtanx$ ?

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