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$$\int_{π/3}^{π/4}\frac{{\sin x\cos x}}{{1 - 2\cos (2x)}}dx= - \frac{{\log 2}}{8}$$ It tells me to prove this. I inputed both and they both equale to -0.0866. Is this enough to prove it or do I have to make the left side equaled to the right- and if so how do you do this? Any help is much appricated. Thank you!

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    $\begingroup$ This is not enough. With that argument I could state that the integral is in fact -866/10000. $\endgroup$
    – Gary
    Mar 4, 2021 at 7:53
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    $\begingroup$ Just treat it like a normal integral - you will get from the LHS to the RHS $\endgroup$
    – user71207
    Mar 4, 2021 at 7:58

2 Answers 2

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Let's consider that: $\sin{x}\cos{x}=\frac{1}{2}\sin{2x}$ and $(-\cos{2x})'=2\sin{2x}$. So: $$\int_{π/3}^{π/4}\frac{{\sin x\cos x}}{{1 - 2\cos (2x)}}dx= \int_{π/3}^{π/4}\frac{{\frac{1}{2}\sin{2x}}}{{1 - 2\cos (2x)}}dx=\frac{1}{8}\log{(1-2\cos{2x}})\lvert_{\pi/3}^{\pi/4}=0-\frac{\log{2}}{8}$$

$\textbf{EDIT:}$ to be more clear let's observe that,

You should have at the numerator the derivative of $(1-2\cos{2x})$ to be able to solve the integral with the logarithm as I have shown...but the numerator it is not exactly the derivative of the denominator, why? $$(1-2\cos{2x})'=(-2\cos{2x})'=-2\cdot(\cos{2x})'=4\sin{2x}$$ But at the numerator we have $\frac{1}{2}\sin{2x}$ so in effect we can multiply and divide by 4 to obtain the desidered form, in this following way: $$\int_{π/3}^{π/4}\frac{{\frac{1}{2}\sin{2x}}}{{1 - 2\cos (2x)}}dx=\int_{π/3}^{π/4}\frac{{\frac{\color{red}4}{\color{red}4 \cdot 2}\sin{2x}}}{{1 - 2\cos (2x)}}dx=$$ $$=\frac{1}{8}\int_{π/3}^{π/4}\frac{{4\sin{2x}}}{{1 - 2\cos (2x)}}dx=\Big[\frac{1}{8}\log{(1-2\cos{2x})}\Big]\Big\lvert_{\pi/3}^{\pi/4}$$

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  • $\begingroup$ where did the 1/8 come from? $\endgroup$
    – user890436
    Mar 4, 2021 at 9:40
  • $\begingroup$ @Emily I have edited my answer with the explanation! Tell me if now it is clear $\endgroup$
    – pawel
    Mar 4, 2021 at 10:12
  • $\begingroup$ i took my time to look at this question and you made it so clear thank you so much you have no idea how much you have helped me.:) $\endgroup$
    – user890436
    Mar 6, 2021 at 11:14
  • $\begingroup$ @Emily You're welcome I am happy that now it is all clear! :) $\endgroup$
    – pawel
    Mar 6, 2021 at 11:18
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So, just calculate this integral:

  1. notice that $\sin(x)\cos(x) = \frac{1}{2}\sin(2x)$
  2. in your integral put $\cos(2x) = t$ then $-2\sin(2x)dx = dt$ What will we have?
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