How do I find the antiderivative of #f(x)=tan(2x) + tan(4x)#?

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Answer 1 · 2018-03-04T18:18:19

#-1/2ln|(cos2x)|-1/4ln|(cos4x)|+c#

or

#1/2ln|sec2x|+1/4ln|sec4x|+c#

the anti derivative is another way of saying integration

so the anti derivative of #f(x)=tan2x+tan4x#

#=int(tan2x+tan4x)dx---(1)#

we will use teh relationship

#int(f'(x))/(f(x))dx=ln|f(x)|+c#

#(1)rarrint((sin2x)/(cos2x)+(sin4x)/(cos4x))dx--(2)#

now

#d/dx(cosnx)=-nsinnx#

so #(2)rarr=-1/2ln|(cos2x)|-1/4ln|(cos4x)|+c--(3)#

#because -ln|cosx|=ln(1/|(cosx)|)=ln|secx|#

#(3)rarr=1/2ln|sec2x|+1/4ln|sec4x|+c#