Need help with this tricky integration question

And yeah god cleaning it up at the end is a nightmare too. Do you want me to work through it again in a clear way then post a picture to get past the integrating stage then you can see if you can clean it up to get to that point.

I understand how the denominator becomes $\displaystyle {cos(2x)}$ but I'm still figuring out the numerator because I have $\displaystyle {sin^{2}(x)-sin(x)cos(x)}$

At this point it is easiest if you split the fraction. You should be able to deal with the sin^2(x) bit using cos(2x)=1-2sin^2(x). You can then use sin(2x)=2sin(x)cos(x) for the second fraction. After that point you should be able to cancel stuff so you have standard integrals. I guess it's still quite difficult if you don't have a formula book on you with a bunch of standard integrals in it. Where is this question from because it sure as hell ain't a c4 question XD.

Oh lol no it's not A Level, it's a problem solving sheet from my university

Do the problem solving sheets count for any of your grade or are they just for supervision type thingies?

Nah it's just practice, we have 1 hour each week taught by a PhD student and we get as far as we can with the work sheet, same goes for my other modules.

So the question is to integrate

$\displaystyle \int{\frac{sin(x)}{sin(x)+cos(x)}}dx$

My working out is below
Note: After I used a second substitution, I realised I was going no one with it and the integration calculators online have strange approaches to this question.

Where does the $\frac{1}{2}$ come in?

I'm left with $-ln|sin(x)+cos(x)|+x+C$ whereas the answer is $-\frac{1}{2}ln|\displaystyle {sin(x)+cos(x)}|+\frac{x}{2}+C$

You have $I+J + I - J = \boxed{2}I = x -\ln|\sin x + \cos x| + C$.This is probably where you dropped the 1/2.

If you can't recognise what the poster above suggested a tangent half angle substitution would also work, but it is longer.

As this integral is for fun, here's a fun method useful for similar integrals:

Let $\displaystyle I = \int \frac{\sin x}{\sin x + \cos x} \mathrm dx$ and $\displaystyle J = \int \frac{\cos x}{\sin x + \cos x}\mathrm dx$.

Consider $I+J$ and $I-J$ to form a system of equations and solve for $I$ and $J$.

Ah fair fairs. Yeah tbh I wouldn't have know what to do without peaking at the answer. I think the general thought process should be that with trig integrals you are gonna want to use identities and to use identities, most have sin or cos squared so you want to make as much sin or cos squared while getting rid of the cos and sin to the power 1. .

Lol I've given up on this, I shouldn't have spent 2 hours on this question but thanks for your contribution

I know you've given up, but another approach that might make sense for you:

Make the substitution u = x + pi / 4, then

sin x + cos x = $\sqrt{2} \sin u$

and sin x = sin(u-pi/4) = sin u cos pi/4 - cos u sin pi /4 = $\dfrac{1}{\sqrt{2}} (\sin u - \cos u)$

This gives you an integral very similar to what you started with, but with numerator and denominator flipped. Since it is much easier to divide sin u and cos u by sin u than it is to divide sin x by (sin x + cos x), this new integral is pretty straightforward.