How do you find the integral of ${tan}^{2} (x) sec (x) d x$ $tan^2 (x)sec(x) dx$ ?

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How do you find the integral of ${tan}^{2} (x) sec (x) d x$ $tan^2 (x)sec(x) dx$ ?

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Answer 1 · 2015-04-25T16:09:10

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

This can be easily be solved by using substitution. Let tan x =u, so that ${sec}^{2} x$ $sec^2x$ dx=du

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

Answer 2 · 2015-04-26T06:56:22

$\frac{1}{2} sec x tan x - \frac{1}{2} ln (sec x + tan x)$ $1/2 secx tanx -1/2 ln(secx +tanx)$

Rewrite integral as, $\int ({sec}^{2} x - 1) sec x d x$ $int(sec^2x-1)sec xdx$ = $\int {sec}^{3} x d x - \int sec x d x$ $intsec^3xdx-int secx dx$ . Now solve $\int {sec}^{3} x d x$ $int sec^3x dx$ by integrating by parts, $(sec x) {sec}^{2} x$ $(sec x) sec^2 x$

=sec x tanx - $\int sec x tan x tan x d x$ $int secx tanx tanx dx$ = sec x tan x - $\int sec x {tan}^{2} x d x$ $int secx tan^2 xdx$

= secx tanx - $\int sec x ({sec}^{2} x - 1) d x$ $int secx (sec^2x-1)dx$ = secx tanx - $\int {sec}^{3} x d x + \int sec x d x$ $int sec^3x dx + int sec xdx$ . Now transpose ${sec}^{3} x d x$ $sec^3xdx$ to the other side and have

$\int {sec}^{3} x d x$ $intsec^3x dx$ = $\frac{1}{2} sec x tan x + \frac{1}{2} \int sec x d x$ $1/2 secx tan x +1/2 int secx dx$ . Thus it is,

$\int {tan}^{2} x sec x d x = \frac{1}{2} sec x tan x + \frac{1}{2} \int sec x d x - \int sec x d x$ $int tan^2x secx dx= 1/2 secx tan x +1/2 int secx dx -int sec xdx$

= $\frac{1}{2} sec x tan x - \frac{1}{2} \int sec x d x$ $1/2 sec x tan x -1/2int secx dx$

= $\frac{1}{2} sec x tan x - \frac{1}{2} ln (sec x + tan x)$ $1/2 secx tan x-1/2 ln(secx+tanx)$ +C

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How do you find the integral of ${tan}^{2} (x) sec (x) d x$ $tan^2 (x)sec(x) dx$ ?

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