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

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

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

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