$Integration headache, please help.$

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Integration headache, I simply cannot see how come that this online calculator (see link) gives the result it gives (see link or attached picture). Please help, it is driving me crazy.

https://www.wolframalpha.com/input/?i2d=true&i=Integrate%5B-Sin%5Bt%5D*Exp%5B-i*t%5D%2C%7Bt%2C-pi%2C0%7D%5D

Integrals

Robin Hyaku

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Using Euler's formula
\[\int_0^{\pi} \sin t e^{it}dt=\int_0^{\pi} \sin t (\cos t+i \sin t)dt\]\[=\int_0^{\pi} \sin t \cos t dt+i \int_0^{\pi}\sin^2 tdt\]
\[=\int_0^{\pi}\frac{\sin 2t}{2}dt+i\int_0^{\pi}\frac{1-\cos 2t}{2}dt\]
\[=-\frac{\cos 2t}{4} |_0^{\pi}+\frac{i}{2}(t -\frac{\sin 2t }{2})|_0^{\pi}\]
\[=(-\frac{1}{4}+\frac{1}{4})+\frac{i}{2}[(\pi -0)-(0-0)]\]
\[=i\frac{\pi}{2}.\]
Here we have used the identities:
\[\sin 2t =\sin t \cos t, \sin^2 t= \frac{1-\cos t}{2}.\]

Nirenberg

Nirenberg

Leave a comment if you need any clarifications.
Robin Hyaku

The given sin(2t) identity I belive to be wrong, and it can be confusing. But I appreaciate the very fast response. Thank you.
Nirenberg

Oh, it' a typo. Sin 2t =2 sin t cos t. There is no mistake in the solution, I just dropped the 2 in the last line when I wrote the trig identity.

$Integration headache, please help.$

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