HARD INTEGRAL (help)

I do not really need help but i made this integral inspired by this thread earlier

http://www.thestudentroom.co.uk/showthread.php?t=3733947

and wondered if all there are any takers with gonads ...


(good surfers with other's work/ideas need not participate)

Lemma: $\displaystyle \frac{12u - 5}{2u-3} = \frac{6(2u - 3)}{2u - 3} + \frac{18-5}{2u-3} = 6 + \frac{13}{2u-3}$

Divide the numerator and denominator of the integrand by $\cos x$ to get

$\displaystyle \int \frac{12u - 5}{2u-3} \, \mathrm{d}x = \int 6 + \frac{13}{2 \tan x - 3} \, \mathrm{d}x$ (can't be bothered with all the tan x's.

This gets you $\displaystyle 6x + 13\int \frac{\mathrm{d}x}{2 \tan x - 3} + \mathcal{C}_0$

Focusing on the above integral a sub of $u = \tan x$ looks promising (I'm doing this as I type. )

$\displaystyle \mathrm{d}u = \sec^2 x = 1 + \tan^2 x = 1 + u^2$

So $\displaystyle \int \frac{13}{2u - 3} \, \frac{\mathrm{dx}}{{\mathrm{d}u}} \, \mathrm{d}u = \int \frac{13}{(2u -3)(1+u^2)}$ (using W.A to do the partial fractions for me, can't be bothered typing them out and they're fairly easy to integrate upon recognition)

$\displaystyle \int \frac{13}{(2u-3)(1+u^2)} \, \mathrm{d}u = 2 \ln (2u -3) - \ln (u^2 + 1) - 3 \arctan u + \mathcal{C}_1$

Putting everything together, we get $\displaystyle 6x -3x + 2 \ln (2 \tan x - 3) - \ln (\sec^2 x) = 3x + 2 \ln (3\cos x - 2 \sin x) + \mathcal{C}_2$

That is the correct answer but there is a much simpler method

I'm sure there is, my way was just brute forcing it after being so mathematically tired out, it's 2 am.

Can you give me any hints for the nice way? Is it considering two different integrals and adding/subtracting?

(by the way I got married in Mauritious in the Grand Gaube Hotel if it still exists, back in 1992)

Can you give me any hints for the nice way? Is it considering two different integrals and adding/subtracting?

even for the adding and subtracting this is not a natural method unless you are given a clue (in my opinion) such as asking for both integrals

this is how I do the integral in the other thread

this might give you more clues

I can almost see it, you get $\int 3 + \frac{2 \cos x - 3\sin x}{3 \cos x + 2 \sin x} \, \mathrm{d}x$, right?

indeed

Great, thanks!

There's another equation amenable to the same trick, something with sech x but I can't recall it right now and I'm heading to bed. :-)