# Prove that $\int_{π/3}^{π/4}\frac{{\sin x\cos x}}{{1 - 2\cos (2x)}}dx= - \frac{{\log 2}}{8}$

$$\int_{π/3}^{π/4}\frac{{\sin x\cos x}}{{1 - 2\cos (2x)}}dx= - \frac{{\log 2}}{8}$$ 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!

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

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: $$\int_{π/3}^{π/4}\frac{{\frac{1}{2}\sin{2x}}}{{1 - 2\cos (2x)}}dx=\int_{π/3}^{π/4}\frac{{\frac{\color{red}4}{\color{red}4 \cdot 2}\sin{2x}}}{{1 - 2\cos (2x)}}dx=$$ $$=\frac{1}{8}\int_{π/3}^{π/4}\frac{{4\sin{2x}}}{{1 - 2\cos (2x)}}dx=\Big[\frac{1}{8}\log{(1-2\cos{2x})}\Big]\Big\lvert_{\pi/3}^{\pi/4}$$

• where did the 1/8 come from?
– user890436
Mar 4, 2021 at 9:40
• @Emily I have edited my answer with the explanation! Tell me if now it is clear Mar 4, 2021 at 10:12
• 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.:)
– user890436
Mar 6, 2021 at 11:14
• @Emily You're welcome I am happy that now it is all clear! :) Mar 6, 2021 at 11:18

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?