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## Main Question or Discussion Point

I have been going through my textbook deriving equations in preparation for my test on QM tomorrow. I noticed in the infinite square well that i was unable to complete the normalization.

My textbook, Griffiths reads :

(integral from 0 to a) ∫|A|^2 * (sin(kx))^2 =|A|^2 * (a/2) =1 Therefore, |A|^2 = 2/a

When i do this integral i do not get this result, furthermore when i put this integral into wolfram alpha i get this result:

http://www.wolframalpha.com/input/?i=integrate+sin^2(kx)+dx+from+0+to+a

Which is also not the required result.

It's probably something stupid but help would be appreciated none the less,

Thanks in advance, John.

My textbook, Griffiths reads :

(integral from 0 to a) ∫|A|^2 * (sin(kx))^2 =|A|^2 * (a/2) =1 Therefore, |A|^2 = 2/a

When i do this integral i do not get this result, furthermore when i put this integral into wolfram alpha i get this result:

http://www.wolframalpha.com/input/?i=integrate+sin^2(kx)+dx+from+0+to+a

Which is also not the required result.

It's probably something stupid but help would be appreciated none the less,

Thanks in advance, John.