How to solve this integral $I = \int\dfrac{\cos^3x}{\sin x + \cos x}dx$? $\displaystyle\int\dfrac{\cos^3x}{\sin x + \cos x}dx$
I added $J =\displaystyle \int\dfrac{\sin^3x}{\sin x + \cos x}dx$
then $I + J = \displaystyle\int\dfrac{\cos^3x + \sin^3x}{\sin x + \cos x}dx = x + \dfrac{1}{2}\cos2x + C$
but I can't find how to solve $I-J$
And is that the true way to solve it?
Please help!

$$I-J = \int \frac{ \cos^3 x - \sin^3 x}{ \sin x  + \cos x} dx =  \int \frac{ (\cos x - \sin x)( \cos^2 x + \sin^2 x + \sin x \cos x )}{ \sin x + \cos x} dx$$
Substitute $$ \sin x + \cos x = t$$
$$t^2 = 1 - 2 \sin x \cos x$$
Or,
$$ \sin x \cos x = \frac{1-t^2}{2}$$
$$ I-J= \int \frac{(1+ ( \frac{1-t^2}{2}))}{t} dt$$
Can you finish?
