How do you integrate ${tan}^{3} x$ $tan^3x$ ?

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Answer 1 · 2017-02-12T15:06:13

$\frac{1}{2} {tan}^{2} x + ln | sec x | + C$ $1/2tan^2x+ln|secx|+C$

by the use of trig identities alternative equivalent solns can be found

$\int {tan}^{3} x d x$ $inttan^3xdx$

now $1 + {tan}^{2} x = {sec}^{2} x = {tan}^{2} x = {sec}^{2} x - 1$ $" "1+tan^2x=sec^2x=tan^2x=sec^2x-1$

$\int {tan}^{3} x d x = \int {tan}^{2} x tan x d x$ $inttan^3xdx=inttan^2xtanxdx$

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

$= \int ({sec}^{2} x tan x) d x - \int tan x d x$ $=int(sec^2xtanx)dx-inttanxdx$

now

$\frac{d}{d x} ({tan}^{2} x) = 2 tan x {sec}^{2} x$ $d/(dx)(tan^2x)=2tanxsec^2x$

$:.=int(sec^2xtanx)dx=1/2tan^2x$

$inttanxdx=intsinx/cosxdx=-ln|cosx|=ln|secx|$

so the complete soln

$inttan^3xdx$

$=1/2tan^2x+ln|secx|+C$

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