Integral

[123]

$\displaystyle \int_{-\pi}^{2\pi}sin\frac{x}{2}dx=...$

 

[123]

Answer is 2, but i miss something, when I solve it my answer is wrong.

 

[soroban]

Hello, 123!

$\displaystyle \int_{\text{-}\pi}^{2\pi}\!\!\sin\frac{x}{2}\,dx$

$\displaystyle \int^{2\pi}_{\text{-}\pi}\!\!\sin\frac{x}{2}\,dx \;=\;\text{-}2\cos\frac{x}{2}\bigg]^{2\pi}_{\text{-}\pi} \;=\;\bigg[\text{-}2\cos\pi\bigg] - \bigg[\text{-}2\cos\left(\text{-}\tfrac{\pi}{2}\right)\bigg] \;=\;(\text{-}2)(\text{-}1) - (\text{-}2)(0) \;=\;2$

 

[123]

$\displaystyle \int^{2\pi}_{\text{-}\pi}\!\!\sin\frac{x}{2}\,dx \;=\;\text{-}2\cos\frac{x}{2}\bigg]^{2\pi}_{\text{-}\pi} \;=$.....

Thanks, but why $\displaystyle -2cos$?

 

[galactus]

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.

 

[123]

now it's clear to me.