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 $Ax+Blog|cosx+sinx|+c$
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.