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

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Answer 1 · 2015-04-25T16:09:10

$tan^3x /3$ +C

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

$int tan^2x sec x dx$ = $intu^2du$ = $u^3/3$ +C= $tan^3x /3$ +C

Answer 2 · 2015-04-26T06:56:22

$1/2 secx tanx -1/2 ln(secx +tanx)$

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

=sec x tanx - $int secx tanx tanx dx$ = sec x tan x - $int secx tan^2 xdx$

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

$intsec^3x dx$ = $1/2 secx tan x +1/2 int secx dx$ . Thus it is,

$int tan^2x secx dx= 1/2 secx tan x +1/2 int secx dx -int sec xdx$

= $1/2 sec x tan x -1/2int secx dx$

= $1/2 secx tan x-1/2 ln(secx+tanx)$ +C

How do you find the integral of tan^2 (x)sec(x) dxtan^2 (x)sec(x) dx?