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L(sin(2t)e^5t)=2/((s-5)^2+2^2)
\[2 \over (s-5)^2+4\]

thanks

any luck?

I guess.. just gimme a minute

thank you so much for your help

then you can substitute the new result in the previous one to get the final integration.

yep, that's what I got as well

now just substitute (2) in (1) to, and take the part with integration in one side..

try it out and see if it works with you

are you talking about the substitution in the end of infinity and zero?