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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=−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)

2πx=udx=duI=02πln(sinu+1+sin2u)du=02πln(1sinu+1+sin2u)du=II=12

Commented by mnjuly1970 last updated on 02/Apr/23

  I=0

I=0

Commented by mehdee42 last updated on 02/Apr/23

yes

yes

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