# How do you integrate ∫ (tan2x + tan4x) dx?

Mar 9, 2015

Re-write it and use substitution.

$\int \left(\tan 2 x + \tan 4 x\right) \mathrm{dx} = \int \left(\frac{\sin 2 x}{\cos 2 x} + \frac{\sin 4 x}{\cos 4 x}\right) \mathrm{dx}$.

Now do the integrals seperately:

$\int \frac{\sin 2 x}{\cos 2 x} \mathrm{dx}$.

Let $u = \cos 2 x$. this makes $\mathrm{du} = - 2 \sin 2 x \mathrm{dx}$, so $\sin 2 x \mathrm{dx} = - \frac{1}{2} \mathrm{du}$.

$\int \frac{\sin 2 x}{\cos 2 x} \mathrm{dx} = - \frac{1}{2} \int \frac{1}{\cos 2 x} \sin 2 x \mathrm{dx} = - \frac{1}{2} \int \frac{1}{u} \mathrm{du}$

So, $\int \frac{\sin 2 x}{\cos 2 x} \mathrm{dx} = - \frac{1}{2} \ln \left\mid \cos 2 x \right\mid$.

In a similar way the second integral is found to be

$\int \frac{\sin 4 x}{\cos 4 x} \mathrm{dx} = - \frac{1}{4} \ln \left\mid \cos 4 x \right\mid$.

So,
$\int \left(\tan 2 x + \tan 4 x\right) \mathrm{dx} = - \frac{1}{2} \ln \left\mid \cos 2 x \right\mid - \frac{1}{4} \ln \left\mid \cos 4 x \right\mid + C$.

Of course, there are many way of rewriting this expression..