f(x)=sin(e^x)-sin[e^(-x)] f(-x)=sin[e^(-x)]-sin(e^x) f(-x)=-{sin(e^x)-sin[e^(-x)]} f(-x)=-f(x) a=ln(pi) b=-a=-ln(pi) An integral of an odd function defined on a symmetrical interval is equal to 0.
@owl3math
14 күн бұрын
Thanks
@taa347
6 ай бұрын
This is my favorite thing about higher math. It’s not so much just figuring the answer but discovering a way to answer
@owl3math
6 ай бұрын
right! The good methods will usually lead to a solution :)
@Ni999
6 ай бұрын
Yep! 😀👍 Mine was similar, after the usub I split it into two integrals and then did the tsub on the 1/u version only. Algebra happened and then it left sinu/u - sinu/u. Celebrated by playing Who Are You. 🎸🎶
@owl3math
6 ай бұрын
ha! Good choice : ) thanks!
@Ni999
6 ай бұрын
@@owl3math Welcome 😃 And thanks for the problem! 👍👍
@sylvainm-v7919
3 ай бұрын
seigneur
@blackroses6315
6 ай бұрын
guessed and got it right, 4 minutes and 28 seconds saved
@owl3math
6 ай бұрын
ha! good move and also useful in a timed contest :)
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