Question

Integrate: (tan x/3 + cot x/3)^2 dx

Answer 1

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Answer 2

This is quite a fun problem, at least the way in which I figured it out:

First, I just squared it:
\[\int (\tan \tfrac{x}{3} + \cot \tfrac{x}{3})^2dx\]
\[\int \tan^2 \tfrac{x}{3} +2+ \cot^2 \tfrac{x}{3} dx\]

notice that the middle term there comes from \(2 \tan \tfrac{x}{3} \cot \tfrac{x}{3} = 2\).

Now from here I recognized I could plug in the Pythagorean identity in two different ways:

\[\tan^2 \tfrac{x}{3} + 1 = \sec^2 \tfrac{x}{3}\]
for example with the other one for cotangent to get:

\[\int \sec^2 \tfrac{x}{3} + \csc^2 \tfrac{x}{3} dx\]

And now you should be able to figure out the rest.