Question: so I've just had my first exam (went pretty well) but I ran into this thing as the first part of the last question.

$$\int \frac{\cos x}{\sin x+\cos x}dx$$

I had a look on wolfram after the exam and it advised to multiply top and bottom by $\sec^3x$. 

Is there another way to tackle this if you didn't know that trick? I find it hard to believe I was meant to know this and it was disproportionately harder than any type of integration question I've come across when practicing.

I couldn't get anywhere when trying to solve this.

Thanks.

Answer: 
Hint: $\sin(x)+\cos(x)=\sqrt{2}\sin(x+\pi/4)$ Then substitute $u=x+\pi/4$.

A more general approach would be the following. Replace $\cos(x)$ and $\sin(x)$ by Euler's formula.

$$\int \frac{f(\cos(x),\sin(x))}{g(\cos(x),\sin(x))}dx=\int \frac{f(\exp(ix),\exp(-ix))}{g(\exp(ix),\exp(-ix))}dx.$$

Now, substitute $u=\exp(ix) \implies du=iudx$.

$$\int \frac{f(u,u^{-1})}{g(u,u^{-1})}\frac{du}{iu}=\int \frac{f(u)}{g(u)}\frac{du}{iu}.$$

If $f$ and $g$ are simple polynomials the resulting integral can always be solved by partial fractions.
Dividing numerator and denominator by $\cos x$ yields

$$\frac{dx}{\tan x+1}$$

Now substitute $x=\tan^{-1}t,dx=\dfrac{dt}{1+t^2}$ to get

$$\int\frac{dt}{(t+1)(t^2+1)}$$

which can now be solved with partial fractions. Similar to the other method but hopefully easier to see.