I know how to solve integration by parts of some functions, but I came across the integration of $-x\sec(2x)$, and I can't solve it. I put the question on an online calculator. I got the answer, but I can't understand the logic behind solving it. Why it used hyperbolic function. I've never solved any question like this before. Isn't there any other way to solve it? What is the logic behind all the substitutions?
How would you integrate $-x\sec(2x)$?
Here is a rough sketch (without checks, tracking branches of log etc).
Assuming we can use integration by parts and pulling minus outside the integral: $$\int x \sec{\left(2 x \right)}\, dx = x \int \sec{\left(2 x \right)}\, dx - \iint \sec{\left(2 x \right)}\, dx\, dx$$ tansforming $\sec(2x)$ to function of $\tan$ and using $\tan(x) = t$ substitution $$ \int \sec{\left(2 x \right)}\, dx = \int \frac{1 + \tan^{2}{\left(x \right)} }{1 - \tan^{2}{\left(x \right)}}\, dx = \int \frac{1}{1 - t^{2}}\, dt = \frac{- \log{\left(x - 1 \right)} + \log{\left(x + 1 \right)}}{2} $$ so after substituting back to first integral $$ \int x \sec{\left(2 x \right)}\, dx = \frac{x \left(- \log{\left(\tan{\left(x \right)} - 1 \right)} + \log{\left(\tan{\left(x \right)} + 1 \right)}\right)}{2} - \int \frac{- \log{\left(\tan{\left(x \right)} - 1 \right)} + \log{\left(\tan{\left(x \right)} + 1 \right)}}{2}\, dx. $$
Considering the second integral, without the $\frac{1}{2}$ factor, there are two integrals to solve: $$ - \int \log{\left(\tan{\left(x \right)} - 1 \right)}\, dx $$ and $$ \int \log{\left(\tan{\left(x \right)} + 1 \right)}\, dx $$ which look quite complicated, but they at least give the idea where does the dilogarithms come from.
Alternative forms of $ \int \sec{\left(2 x \right)}\, dx $ that could be tried: $$\frac{\log{\left(\tan{\left(2 x \right)} + \sec{\left(2 x \right)} \right)}}{2}$$ and $$ \frac{- \log{\left(\sin{\left(2 x \right)} - 1 \right)} + \log{\left(\sin{\left(2 x \right)} + 1 \right)}}{4}$$ or piecewise acoth(t) and atanh(t) depending on value of $t$, if hyperbolic functions are wanted. Finally more care is requred to solve it generally and not loose track of the branch of logarithm, check what conditions on $x$ do these transformations impose etc.
Anyway, the resulting integrals (that would have to be integrated over dx after the integration by parts) aren't really much easier (such as $\log(\sin(2x) + 1), \log(\cos(2x)), \log(\tan(x))$ etc.). Surprisingly, Integral Calculator can do some of them step-by-step, so it can be followed how to get them.