How do you find the antiderivative of $\int {tan}^{3} x sec x d x$ $int tan^3xsecx dx$ ?

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How do you find the antiderivative of $\int {tan}^{3} x sec x d x$ $int tan^3xsecx dx$ ?

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Answer 1 · 2017-02-02T23:16:02

$\int {tan}^{3} x sec x d x = \frac{1}{3} {sec}^{3} x - sec x + C$ $inttan^3xsecxdx=1/3sec^3x-secx+C$

Explanation:

$I = \int {tan}^{3} x sec x d x$ $I=inttan^3xsecxdx$

Rewrite this using ${tan}^{2} x + 1 = {sec}^{2} x$ $tan^2x+1=sec^2x$ :

$I = \int {tan}^{2} x tan x sec x d x$ $I=inttan^2xtanxsecxdx$

$I = \int ({sec}^{2} x - 1) (sec x tan x) d x$ $I=int(sec^2x-1)(secxtanx)dx$

Let $u = sec x$ $u=secx$ . Then $d u = sec x tan x d x$ $du=secxtanxdx$ :

$I = \int (u^{2} - 1) d u$ $I=int(u^2-1)du$

$I = \frac{1}{3} u^{3} - u + C$ $I=1/3u^3-u+C$

$I = \frac{1}{3} {sec}^{3} x - sec x + C$ $I=1/3sec^3x-secx+C$

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How do you find the antiderivative of $\int {tan}^{3} x sec x d x$ $int tan^3xsecx dx$ ?

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Explanation:

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How do you find the antiderivative of int tan^3xsecx dx∫tan3xsecxdxint tan^3xsecx dx?

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How do you find the antiderivative of $\int {tan}^{3} x sec x d x$ $int tan^3xsecx dx$ ?