How do you integrate #int sinx/cos^3xdx#?
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1 Answer
The derivative of
Explanation:
# = -int u^-3 du#
# = -u^-2/(-2) +C = 1/(2u^2) +C#
#=1/(2cos^2 x) +C#
#=1/2sec^2x + C#
(Check the answer by differentiation.)
