How do you integrate $tan x \cdot {sec}^{3} x d x$ $tanx * sec^3x dx$ ?

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How do you integrate $tan x \cdot {sec}^{3} x d x$ $tanx * sec^3x dx$ ?

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Answer 1 · 2017-01-15T15:46:35

$\frac{1}{3} {sec}^{3} x + C$ $1/3sec^3x+C$

Explanation:

We have:

$\int tan x {sec}^{3} x d x$ $inttanxsec^3xdx$

Notice that we can write this in terms of $sec x$ $secx$ and the derivative of $sec x$ $secx$ , which is $sec x tan x$ $secxtanx$ .

Make the substitution $u = sec x$ $u=secx$ , implying that $d u = sec x tan x d x$ $du=secxtanxdx$ .

We can rewrite the integral as such:

$= \int {sec}^{2} x (sec x tan x d x) = \int u^{2} d u = \frac{1}{3} u^{3} + C = \frac{1}{3} {sec}^{3} x + C$ $=intsec^2x(secxtanxdx)=intu^2du=1/3u^3+C=1/3sec^3x+C$

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How do you integrate $tan x \cdot {sec}^{3} x d x$ $tanx * sec^3x dx$ ?

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How do you integrate $tan x \cdot {sec}^{3} x d x$ $tanx * sec^3x dx$ ?