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.
HINT:

For $I=\displaystyle\int\dfrac{a\cos x+b\sin x}{c\cos x+d\sin x}dx,$

write $a\cos x+b\sin x=A(c\cos x+d\sin x)+B\cdot\dfrac{d(c\cos x+d\sin x)}{dx}$ where $A,B$ are arbitrary constants so that

$I=\displaystyle A\int\ dx+B\int\dfrac{d(c\cos x+d\sin x)}{c\cos x+d\sin x}$

Compare the coefficients of $\cos x,\sin x$ to find $A,B$