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How do you find $\int 1 - {tan}^{2} x d x$ $int 1-tan^2xdx$ ?

1 Answer

Nov 15, 2015

Use trigonometric identity: ${tan}^{2} x = {sec}^{2} x - 1$ $tan^2x = sec^2x-1$ to rewrite.

Explanation:

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

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

$= 2 x - tan x + C$ $= 2x-tanx +C$

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