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Question Number 139548 by peter frank last updated on 28/Apr/21
Answered by Dwaipayan Shikari last updated on 28/Apr/21
∫−π2π2x2sin(x)1+x6dx=−∫−π2π2x2sin(x)1+x6dx=I2I=0⇒I=0
Commented by peter frank last updated on 29/Apr/21
thankyou
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