Question Number 190385 by TUN last updated on 02/Apr/23
Answered by mehdee42 last updated on 02/Apr/23
$$\mathrm{2}\pi−{x}={u}\Rightarrow{dx}=−{du} \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:{ln}\left(−{sinu}+\sqrt{\mathrm{1}+{sin}^{\mathrm{2}} {u}}\right){du} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:{ln}\left(\frac{\mathrm{1}}{{sinu}+\sqrt{\mathrm{1}+{sin}^{\mathrm{2}} {u}}}\right){du}=−{I} \\ $$$$\Rightarrow{I}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by mnjuly1970 last updated on 02/Apr/23
$$\:\:\mathrm{I}=\mathrm{0} \\ $$
Commented by mehdee42 last updated on 02/Apr/23
$${yes} \\ $$