How do I evaluate the indefinite integral #inttan^2(x)dx# ?
1 Answer
Jul 30, 2014
#=tanx-x+c# , where#c# is a constantUsing Trigonometric Identity, which is
#sec^2x-tan^2x=1#
#tan^2x=sec^2x-1# Using this Trigonometric Identity in integration,
#=int(sec^2x-1)dx#
#=intsec^2xdx-intdx#
#=tanx-x+c# , where#c# is a constant
