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Evaluate $$\int \frac{dx}{\sin{x}-\cos{x}} $$
I know it can be done by Weierstrass substitution. But I am looking for new/simple approach. For example I tried:
$$\int \frac{1}{\sin{x}-\cos{x}} \cdot \frac{\sin{x}+\cos{x}}{\sin{x}+\cos{x}}dx=\int \frac{\sin{x}+\cos{x}}{-\cos{2x}}dx ,$$ but I can't continue from here.
Write $$\sin{x}-\cos{x}=\sqrt2\left(\frac1{\sqrt2}\sin{x}-\frac1{\sqrt2}\cos{x}\right)=\sqrt2\sin\left(x-\frac{\pi}4\right).$$
Hint Using an angle sum formula gives $$\sin x - \cos x = \sqrt{2} \sin \left(x - \frac{\pi}{4}\right) .$$
Multiply and divide the denominator by $\frac{\sqrt{2}}{2}$ which is equal to $\sin(\frac{\pi}{4})=\cos(\frac{\pi}{4})$
Now use the fact that $\sin(u-v)=\sin(u)\cos(v)-\cos(u)\sin(v)$.
A continuation to your work
$$\frac1{\sin x-\cos x}=\frac{\sin x+\cos x}{\sin^2x-\cos^2x}=\frac{\sin x}{1-2\cos^2x}+\frac{\cos x}{2\sin^2-1}$$
Then
$$I=-\frac1{\sqrt{2}}\tanh^{-1}(\sqrt{2}\cos x)-\frac1{\sqrt{2}}\tanh^{-1}(\sqrt{2}\sin x)+C$$
