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Question Number 190385 by TUN last updated on 02/Apr/23
Answered by mehdee42 last updated on 02/Apr/23
2π−x=u⇒dx=−duI=∫02πln(−sinu+1+sin2u)du=∫02πln(1sinu+1+sin2u)du=−I⇒I=12
Commented by mnjuly1970 last updated on 02/Apr/23
I=0
Commented by mehdee42 last updated on 02/Apr/23
yes
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