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: 
**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 $Ax+Blog|cosx+sinx|+c$
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 $A,B$