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I want to solve the following integral:

$$\int \frac{dx}{\sqrt{1-\sin(2x)}}$$

My attempt: $$\int \frac{dx}{\sqrt{1-\sin(2x)}}$$ $$=\int \frac{dx}{\sqrt{(\sin x- \cos x)^2}}\tag{1}$$ $$=\int \frac{dx}{\sin x- \cos x} \qquad \tag{2}$$

After this, I used this method and got the final answer as, $$\displaystyle \dfrac1{\sqrt{2}} \ln \left|\csc\left(x - \dfrac\pi4\right) - \cot\left(x - \dfrac\pi4\right)\right|+ C$$

But, the problem is that WolframAlpha returns answer as an imaginary solution. I want to know how can we approach this complex solution.

Also I'm not sure if moving from $(1)$ to $(2)$ is even valid; as I'm ignoring the negative version.

  • Yes, I try this and same. The assumption command doesn't work for the indefinite integral case. – MathFail Jul 28 '22 at 18:58
  • You can use directly $1-\sin(2x)=2\sin(x-\frac{\pi}4)^2$ and also since the integral is indefinite, just ignore the offset $-\frac{\pi}4$ as well to get to $\int\frac{dx}{\sqrt{2}\sin(x)}$. – zwim Jul 28 '22 at 20:06
  • Have you tried using $\sin(\theta) = \frac{-i}{2}(e^{i\theta} - e^{-i\theta})$? That may be where WA is getting the complex numbers from. – Dan Jul 28 '22 at 20:33
  • The first solution W.A. gives for me is actually real: For the principal branch of $\sqrt[4]{\cdot}$, the complex coefficient $(1-i) \sqrt[4]{-1}$ is just $(1 - i) \cdot \frac{1 + i}{\sqrt{2}} = \sqrt{2}$. I'm not sure why W.A. doesn't simplify this expression. – Travis Willse Jul 29 '22 at 07:47

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