1)
$$\int\frac{1}{\sin(x)+\cos(x)}\,dx=\int\frac{1}{\sqrt2\sin\left(x+\frac{\pi}{4}\right)}\,dx$$
$$=-\frac{1}{\sqrt2}\ln\left|\cot\left(x+\frac{\pi}{4}\right)+\csc\left(x+\frac{\pi}{4}\right)\right|+C$$
2) Unfortunately, this method of solution introduces a zero where the original integrand is defined. As an indefinite integral it is correct in the sense that its derivative is equivalent to the integrand, under certain circumstances.
$$\int\frac{1}{\cos(x)+\sin(x)}\,dx=\int\frac{\cos(x)-\sin(x)}{\cos(2x)}\,dx$$
$$=\int\frac{\cos(x)}{1-2\sin^2(x)}\,dx+\int\frac{\sin(x)}{1-2\cos^2(x)}\,dx$$
$$=\frac{1}{2\sqrt2}\left(\ln\left|1+\sqrt2\sin(x)\right|-\ln\left|1-\sqrt2\sin(x)\right|\right)$$
$$+\frac{1}{2\sqrt2}\left(\ln\left|1-\sqrt2\cos(x)\right|-\ln\left|1+\sqrt2\cos(x)\right|\right)+C$$
3)
$$\int\frac{1}{\sin(x)+\cos(x)}\,dx=\int\frac{1}{\frac{e^{ix}+e^{-ix}}{2}+\frac{e^{ix}-e^{-ix}}{2i}}\,dx$$
$$=\int\frac{2i}{(i+1)e^{ix}+(i-1)e^{-ix}}\,dx=\int\frac{2ie^{ix}}{(i+1)e^{2ix}+(i-1)}\,dx$$
$$\sqrt i\tan w=e^{ix},\quad\sqrt i\sec^2w\,dw=ie^{ix}\,dx$$
$$\sqrt2i\int\,dw=\sqrt2i w+C=-\sqrt2i\arctan\left(\frac{e^{ix}}{\sqrt i}\right)+C$$
$$=-\sqrt2i\arctan\left(e^{i(x-\pi/4)}\right)+C$$
(we need a negative to account for $i^2=-1$).
