How do you integrate ${(tan x)}^{8}$ $(tanx)^8$ ?

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How do you integrate ${(tan x)}^{8}$ $(tanx)^8$ ?

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Answer 1 · 2017-04-13T10:25:39

$\frac{{tan}^{7} x}{7} - \frac{{tan}^{5} x}{5} + \frac{{tan}^{3} x}{3} - tan x + x + C .$ $tan^7x/7-tan^5x/5+tan^3x/3-tanx+x+C.$

Explanation:

Let, $I = \int {tan}^{8} x d x = \int {tan}^{6} x ({sec}^{2} x - 1) d,$ $I=inttan^8xdx=inttan^6x(sec^2x-1)d,$

$= \int {(tan x)}^{6} d (tan x) - \int {tan}^{4} x ({sec}^{2} x - 1) d x,$ $=int(tanx)^6d(tanx)-inttan^4x(sec^2x-1)dx,$

$= \frac{{tan}^{7} x}{7} - \int {(tan x)}^{4} d (tan x) + \int {tan}^{2} x ({sec}^{2} x - 1) d x,$ $=tan^7x/7-int(tanx)^4d(tanx)+inttan^2x(sec^2x-1)dx,$

$= \frac{{tan}^{7} x}{7} - \frac{{tan}^{5} x}{5} + \int {(tan x)}^{2} d (tan x) - \int {tan}^{2} x d x,$ $=tan^7x/7-tan^5x/5+int(tanx)^2d(tanx)-inttan^2xdx,$

$= \frac{{tan}^{7} x}{7} - \frac{{tan}^{5} x}{5} + \frac{{tan}^{3} x}{3} - \int ({sec}^{2} x - 1) d x .$ $=tan^7x/7-tan^5x/5+tan^3x/3-int(sec^2x-1)dx.$

$:. I=tan^7x/7-tan^5x/5+tan^3x/3-tanx+x+C.$

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How do you integrate ${(tan x)}^{8}$ $(tanx)^8$ ?

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