Definite integral involving Log, Sin and evaluates to Zero

Question

I need to evaluate this integral.

$$\int_{- \frac{\pi}{2}} ^{\frac{\pi}{2}} \ln( |1 + 2 \sin{x} |) ~ \mathrm{d} x$$

By using Wolfram, the answer is 0.

How do I solve this algebraically?

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My progress -

\begin{align} I = \int _{-\frac{\pi}{2}} ^{\frac{\pi}{2}} \ln |1+2 \sin{x}| \, dx &= \int _0 ^{\frac{\pi}{2}} \ln |1-4 \sin ^2 {x}| \, dx \\\\ &= \int _0 ^{\frac{\pi}{4}} \ln |1-4 \sin ^2 {x}| \, dx + \int _{\frac{\pi}{2}} ^{\frac{\pi}{2}} \ln |1-4 \sin ^2 {x}| \, dx \\\\ &= \int _{0} ^{\frac{\pi}{4}} \ln |1 - 2 \sin{2x}| \, dx + \int _{\frac{\pi}{4}} ^{\frac{\pi}{2}} \ln |1+2 \sin{2x}| \, dx \end{align}

The motivation for the last step is to somehow get similar looking integrals so that I can finally get an expression of the type $I = k I$

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Thank you for all the comments, I have edited the post to include my progress.

Answer 1

I have found a proof by an alternate approach. However, if someone could help me complete the approach suggested in the OP, I would be greatful.

\begin{align} I = \int _{-\frac{\pi}{2}} ^{\frac{\pi}{2}} \ln |1+2 \sin{x}| \, dx &= \int _0 ^{\frac{\pi}{2}} \ln |1-4 \sin ^2 {x}| \, dx \\\\ &= \int_0^{\frac{\pi}{2}}\ln|1-2\cos{2x}| \,dx \\\\ &= \int_0^{\pi}\frac{1}{2} \ln|1-2\cos{x}| \, dx \\\\ &= \int_0^{\frac{\pi}{2}} \frac{1}{2} \ln|1-4\cos^2{x}| \, dx \\\\ &= \int _0 ^{\frac{\pi}{2}} \frac{1}{2} \ln |1-4 \sin ^2 {x}| \, dx \\\\ &= \frac{I}{2} \end{align}

Hence, we get $I = 0$

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I think I also need to prove it's convergence as the integrand goes to $- \infty$ also. But I don't know how to do that.