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Question-190385




Question Number 190385 by TUN last updated on 02/Apr/23
Answered by mehdee42 last updated on 02/Apr/23
2π−x=u⇒dx=−du  I=∫_0 ^(2π)  ln(−sinu+(√(1+sin^2 u)))du  =∫_0 ^(2π)  ln((1/(sinu+(√(1+sin^2 u)))))du=−I  ⇒I=(1/2)
$$\mathrm{2}\pi−{x}={u}\Rightarrow{dx}=−{du} \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:{ln}\left(−{sinu}+\sqrt{\mathrm{1}+{sin}^{\mathrm{2}} {u}}\right){du} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:{ln}\left(\frac{\mathrm{1}}{{sinu}+\sqrt{\mathrm{1}+{sin}^{\mathrm{2}} {u}}}\right){du}=−{I} \\ $$$$\Rightarrow{I}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by mnjuly1970 last updated on 02/Apr/23
  I=0
$$\:\:\mathrm{I}=\mathrm{0} \\ $$
Commented by mehdee42 last updated on 02/Apr/23
yes
$${yes} \\ $$

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