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Integral
- Thread starter 123
- Start date
Hello, 123!
\(\displaystyle \int^{2\pi}_{\text{-}\pi}\!\!\sin\frac{x}{2}\,dx \;=\;\text{-}2\cos\frac{x}{2}\bigg]^{2\pi}_{\text{-}\pi} \;=\;\bigg[\text{-}2\cos\pi\bigg] - \bigg[\text{-}2\cos\left(\text{-}\tfrac{\pi}{2}\right)\bigg] \;=\;(\text{-}2)(\text{-}1) - (\text{-}2)(0) \;=\;2\)
\(\displaystyle \int_{\text{-}\pi}^{2\pi}\!\!\sin\frac{x}{2}\,dx\)
\(\displaystyle \int^{2\pi}_{\text{-}\pi}\!\!\sin\frac{x}{2}\,dx \;=\;\text{-}2\cos\frac{x}{2}\bigg]^{2\pi}_{\text{-}\pi} \;=\;\bigg[\text{-}2\cos\pi\bigg] - \bigg[\text{-}2\cos\left(\text{-}\tfrac{\pi}{2}\right)\bigg] \;=\;(\text{-}2)(\text{-}1) - (\text{-}2)(0) \;=\;2\)
Thanks, but why \(\displaystyle -2cos\)?soroban said:\(\displaystyle \int^{2\pi}_{\text{-}\pi}\!\!\sin\frac{x}{2}\,dx \;=\;\text{-}2\cos\frac{x}{2}\bigg]^{2\pi}_{\text{-}\pi} \;=\).....
Make the substitution \(\displaystyle u=\frac{x}{2}\)
Then, \(\displaystyle du=\frac{1}{2}dx\Rightarrow 2du=dx\)
That is why.
Upon making the subs we get \(\displaystyle 2\int sin(u)du\)
Now integrate and you get Soroban's result.
Then, \(\displaystyle du=\frac{1}{2}dx\Rightarrow 2du=dx\)
That is why.
Upon making the subs we get \(\displaystyle 2\int sin(u)du\)
Now integrate and you get Soroban's result.