Question Number 227043 by Spillover last updated on 28/Dec/25
Answered by Spillover last updated on 28/Dec/25
Answered by Spillover last updated on 28/Dec/25
Answered by Ghisom_ last updated on 28/Dec/25
![I=∫_0 ^π ln ((sin x)/(1+cos x)) dx= [t=π−x] =∫_0 ^π ln ((sin t)/(1−cos t)) dt ⇒ 2I=∫_0 ^π ln ((sin x)/(1+cos x)) +ln ((sin x)/(1−cos x)) dx= =∫_0 ^π ln ((sin^2 x)/((1+cos x)(1−cos x))) dx= =∫_0 ^π ln 1 dx=∫_0 ^π 0dx=0](https://www.tinkutara.com/question/Q227048.png)
$${I}=\underset{\mathrm{0}} {\overset{\pi} {\int}}\mathrm{ln}\:\frac{\mathrm{sin}\:{x}}{\mathrm{1}+\mathrm{cos}\:{x}}\:{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\pi−{x}\right] \\ $$$$=\underset{\mathrm{0}} {\overset{\pi} {\int}}\mathrm{ln}\:\frac{\mathrm{sin}\:{t}}{\mathrm{1}−\mathrm{cos}\:{t}}\:{dt} \\ $$$$\Rightarrow \\ $$$$\mathrm{2}{I}=\underset{\mathrm{0}} {\overset{\pi} {\int}}\mathrm{ln}\:\frac{\mathrm{sin}\:{x}}{\mathrm{1}+\mathrm{cos}\:{x}}\:+\mathrm{ln}\:\frac{\mathrm{sin}\:{x}}{\mathrm{1}−\mathrm{cos}\:{x}}\:{dx}= \\ $$$$=\underset{\mathrm{0}} {\overset{\pi} {\int}}\mathrm{ln}\:\frac{\mathrm{sin}^{\mathrm{2}} \:{x}}{\left(\mathrm{1}+\mathrm{cos}\:{x}\right)\left(\mathrm{1}−\mathrm{cos}\:{x}\right)}\:{dx}= \\ $$$$=\underset{\mathrm{0}} {\overset{\pi} {\int}}\mathrm{ln}\:\mathrm{1}\:{dx}=\underset{\mathrm{0}} {\overset{\pi} {\int}}\mathrm{0}{dx}=\mathrm{0} \\ $$
Commented by Spillover last updated on 28/Dec/25
$${thanks} \\ $$
Answered by Spillover last updated on 28/Dec/25
