What's the integral of ${(tan x)}^{2}$ $(tanx)^2$ ?

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What's the integral of ${(tan x)}^{2}$ $(tanx)^2$ ?

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Answer 1 · 2016-03-07T23:21:34

It is

$\int {(tan x)}^{2} d x = \int \frac{{sin}^{2} x}{{cos}^{2} x} d x = \int \frac{1 - {cos}^{2} x}{{cos}^{2} x} d x = \int \frac{1}{{cos}^{2} x} d x - \int 1 d x = tan x - x + c$ $int (tanx)^2dx=int (sin^2x)/cos^2xdx=int (1-cos^2x)/cos^2xdx= int 1/(cos^2x)dx-int 1dx=tanx-x+c$

where $c$ $c$ is the integration constant

Note we used the following trigonometric identities

1) $tan x = \frac{sin x}{cos x}$ $tanx=sinx/cosx$

2) ${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x+cos^2x=1$

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