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

*

*notice that $\sin(x)\cos(x) = \frac{1}{2}\sin(2x)$

*in your integral put $\cos(2x) = t$ then $-2\sin(2x)dx = dt$
What will we have?

