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

Mar 4, 2018

$- \frac{1}{2} \ln | \left(\cos 2 x\right) | - \frac{1}{4} \ln | \left(\cos 4 x\right) | + c$

or

$\frac{1}{2} \ln | \sec 2 x | + \frac{1}{4} \ln | \sec 4 x | + c$

Explanation:

the anti derivative is another way of saying integration

so the anti derivative of $f \left(x\right) = \tan 2 x + \tan 4 x$

$= \int \left(\tan 2 x + \tan 4 x\right) \mathrm{dx} - - - \left(1\right)$

we will use teh relationship

$\int \frac{f ' \left(x\right)}{f \left(x\right)} \mathrm{dx} = \ln | f \left(x\right) | + c$

$\left(1\right) \rightarrow \int \left(\frac{\sin 2 x}{\cos 2 x} + \frac{\sin 4 x}{\cos 4 x}\right) \mathrm{dx} - - \left(2\right)$

now

$\frac{d}{\mathrm{dx}} \left(\cos n x\right) = - n \sin n x$

so $\left(2\right) \rightarrow = - \frac{1}{2} \ln | \left(\cos 2 x\right) | - \frac{1}{4} \ln | \left(\cos 4 x\right) | + c - - \left(3\right)$

$\because - \ln | \cos x | = \ln \left(\frac{1}{|} \left(\cos x\right) |\right) = \ln | \sec x |$

$\left(3\right) \rightarrow = \frac{1}{2} \ln | \sec 2 x | + \frac{1}{4} \ln | \sec 4 x | + c$