How do you integrate $\int (tan 2 x + tan 4 x) d x$ $∫ (tan2x + tan4x) dx$ ?

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How do you integrate $\int (tan 2 x + tan 4 x) d x$ $∫ (tan2x + tan4x) dx$ ?

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Answer 1 · 2015-03-09T21:52:40

Re-write it and use substitution.

$\int (tan 2 x + tan 4 x) d x = \int (\frac{sin 2 x}{cos 2 x} + \frac{sin 4 x}{cos 4 x}) d x$ $int(tan2x+tan4x)dx=int((sin2x)/(cos2x)+(sin4x)/(cos4x))dx$ .

Now do the integrals seperately:

$\int \frac{sin 2 x}{cos 2 x} d x$ $int(sin2x)/(cos2x)dx$ .

Let $u = cos 2 x$ $u=cos2x$ . this makes $d u = - 2 sin 2 x d x$ $du=-2sin2x dx$ , so $sin 2 x d x = - \frac{1}{2} d u$ $sin2x dx = -1/2du$ .

$\int \frac{sin 2 x}{cos 2 x} d x = - \frac{1}{2} \int \frac{1}{cos 2 x} sin 2 x d x = - \frac{1}{2} \int \frac{1}{u} d u$ $int(sin2x)/(cos2x)dx=-1/2int1/(cos2x)sin2x dx=-1/2 int 1/udu$

So, $\int \frac{sin 2 x}{cos 2 x} d x = - \frac{1}{2} ln | cos 2 x |$ $int(sin2x)/(cos2x)dx=-1/2 ln abs (cos 2x)$ .

In a similar way the second integral is found to be

$\int \frac{sin 4 x}{cos 4 x} d x = - \frac{1}{4} ln | cos 4 x |$ $int(sin4x)/(cos4x)dx=-1/4 ln abs (cos 4x)$ .

So,
$\int (tan 2 x + tan 4 x) d x = - \frac{1}{2} ln | cos 2 x | - \frac{1}{4} ln | cos 4 x | + C$ $int(tan2x+tan4x)dx=-1/2 ln abs (cos 2x)-1/4 ln abs (cos 4x)+C$ .

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How do you integrate $\int (tan 2 x + tan 4 x) d x$ $∫ (tan2x + tan4x) dx$ ?

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

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How do you integrate $\int (tan 2 x + tan 4 x) d x$ $∫ (tan2x + tan4x) dx$ ?