$\int \frac{dx}{\sin x-\cos x}$

Question

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.

Answer 1

Write $$\sin{x}-\cos{x}=\sqrt2\left(\frac1{\sqrt2}\sin{x}-\frac1{\sqrt2}\cos{x}\right)=\sqrt2\sin\left(x-\frac{\pi}4\right).$$

Answer 2

Hint Using an angle sum formula gives $$\sin x - \cos x = \sqrt{2} \sin \left(x - \frac{\pi}{4}\right) .$$

Answer 3

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 $\sin(u-v)=\sin(u)\cos(v)-\cos(u)\sin(v)$.

Answer 4

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

$$I=-\frac1{\sqrt{2}}\tanh^{-1}(\sqrt{2}\cos x)-\frac1{\sqrt{2}}\tanh^{-1}(\sqrt{2}\sin x)+C$$

Answer 5

Similarly to the other answers, notice that: \begin{align*} \sin(x) - \cos(x) & = \sqrt{2}\left(\sin(x)\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\cos(x)\right)\\\\ & = -\sqrt{2}\left(\cos(x)\cos\left(\frac{\pi}{4}\right) - \sin(x)\sin\left(\frac{\pi}{4}\right)\right)\\\\ & = -\sqrt{2}\cos\left(x + \frac{\pi}{4}\right) \end{align*}

This is a slightly different approach from the answer proposed here.