Recent content by sunveer

I solved it by Integration by parts but at last I am getting (1+a^2){∫e−2xcos(2ax)dx}=0

1. Evaluate [PLAIN]http://www.goiit.com/equations/2011/5/25/795493e8-6a4d-42b0-8c18-2afa4e75b653.png I have tried hard to solve it but I am not getting the exact answer. The ans is given to be -a^2/(1+a^2)

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