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: 
Note that $\cos x = \frac{1}{2} ( 2 \cos x + \sin x - \sin x)$.
So then
$$ \int \frac{\cos x}{\sin x + \cos x} dx = \frac{1}{2} \int \frac{\cos x - \sin x}{\cos x + \sin x}dx + \frac{1}{2}\int \frac{\cos x + \sin x}{\cos x + \sin x} dx.$$
The second is easy to integrate, and the first is now just a $u$-substituteion as the numerator is the derivative of the denominator.
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.