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.
**Method 1:-**$$\frac{cosx}{cosx+sinx}=\frac{cos(x-π/4 +π/4)}{\sqrt {2}cos(x-π/4)}=\frac{cos(x-π/4)-sin(x-π/4)}{2cos(x-π/4)}$$

**Method 2:-** Write the numerator

$cosx=A(cosx+sinx)+B\frac{d}{dx}(cosx+sinx)$

to find $A$ and $B$ then the answer will be $Ax+Blog|cosx+sinx|+c$