How do you find the antiderivative of #int sinx/cos^3x dx#?

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Answer 1 · 2016-12-26T11:11:43

The answer is #=1/(2cos^2x)+C#

We need

#intx^ndx=x^(n+1)/(n+1)+C(x!=0)#

#(cosx)'=-sinx#

We perform this integral by substitution

Let #u=cosx#, #=>#, #du=-sinxdx#

Therefore,

#int(sinxdx)/(cos^3x)=int(-du)/u^3#

#=1/(2u^2)=1/(2cos^2x)+C#