How do I evaluate the indefinite integral $\int {tan}^{2} (x) d x$ $inttan^2(x)dx$ ?

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How do I evaluate the indefinite integral $\int {tan}^{2} (x) d x$ $inttan^2(x)dx$ ?

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Answer 1 · 2014-07-30T17:34:55

$= tan x - x + c$ $=tanx-x+c$ , where $c$ $c$ is a constant

Using Trigonometric Identity, which is

${sec}^{2} x - {tan}^{2} x = 1$ $sec^2x-tan^2x=1$

${tan}^{2} x = {sec}^{2} x - 1$ $tan^2x=sec^2x-1$

Using this Trigonometric Identity in integration,

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

$= \int {sec}^{2} x d x - \int d x$ $=intsec^2xdx-intdx$

$= tan x - x + c$ $=tanx-x+c$ , where $c$ $c$ is a constant

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How do I evaluate the indefinite integral $\int {tan}^{2} (x) d x$ $inttan^2(x)dx$ ?

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How do I evaluate the indefinite integral $\int {tan}^{2} (x) d x$ $inttan^2(x)dx$ ?