How do you integrate ${sec}^{3} x (tan x) d x$ $sec^3x (tanx) dx$ ?

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How do you integrate ${sec}^{3} x (tan x) d x$ $sec^3x (tanx) dx$ ?

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Answer 1 · 2016-10-04T22:37:01

$\frac{{sec}^{3} x}{3} + C$ $sec^3x/3+C$

Explanation:

When working with integrals of secant and tangent, it's important to remember the following:

$\frac{d}{d x} tan x = {sec}^{2} x$ $d/dxtanx=sec^2x$
$\frac{d}{d x} sec x = sec x tan x$ $d/dxsecx=secxtanx$

Here, we see that we can write ${sec}^{3} x (tan x)$ $sec^3x(tanx)$ as ${sec}^{2} x (sec x tan x)$ $sec^2x(secxtanx)$ , which is perfect, since it composed of ${sec}^{2} x$ $sec^2x$ and the derivative of secant, $sec x tan x$ $secxtanx$ . This indicates to us that we want to use a substitution of $u = sec x$ $u=secx$ .

$\int {sec}^{3} x (tan x) d x = \int {sec}^{2} x (sec x tan x) d x$ $intsec^3x(tanx)dx=intsec^2x(secxtanx)dx$

With $u = sec x$ $u=secx$ and $d u = (sec x tan x) d x$ $du=(secxtanx)dx$ :

$= \int u^{2} d u = \frac{u^{3}}{3} + C = \frac{{sec}^{3} x}{3} + C$ $=intu^2du=u^3/3+C=sec^3x/3+C$

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How do you integrate ${sec}^{3} x (tan x) d x$ $sec^3x (tanx) dx$ ?

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