Mistake in evaluating $ \int {\frac{\cos 2x}{(\cos x+\sin x)^2 }}dx$

Question

I know I've got the incorrect answer. Can anyone spot where I did wrong?

Answer 1

$$\dfrac{d(\ln(\cos x+\sin x))}{dx}=\dfrac{\cos x-\sin x}{\cos x+\sin x}=\dfrac{\cos^2x-\sin^2x}{(\cos x+\sin x)^2}$$ if $\cos x+\sin x\ne0$

So, you are correct.

Alternatively, $$I=\int\dfrac{\cos2x\ dx}{(\cos x+\sin x)^2}=\int\dfrac{\cos2x\ dx}{1+\sin2x}$$

Set $1+\sin2x=u,dx=?$

$$2I=\dfrac{du}u=\ln|u|=\ln|(\cos x+\sin x)^2|=2\ln|\cos x+\sin x|+K$$

Answer 2

You should state what your expected "correct" answer is, rather than making us guess.

I'm guessing it's probably $\frac 12 \ln |1 + \sin 2x| + c$, which is completely equivalent to your answer by conversion with a trigonometric identity. The arbitrary constant doesn't even come into play here.