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How do you find #int 1-tan^2xdx#?

1 Answer

Nov 15, 2015

Use trigonometric identity: #tan^2x = sec^2x-1# to rewrite.

Explanation:

#int (1-tan^2x)dx = int(1-(sec^2x-1))dx#

# = int(2-sec^2x) dx#

# = 2x-tanx +C#

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