int_{-frac{pi}{2}}^{frac{pi}{2}} sin (x-[x]) | vedclass

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Question

(A) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin (x-[x]) d x$. Since $[x]$ is the Greatest Integer Function,we split the integral based on the

Vedclass · Accepted Answer

$3(1-\cos 1)+\sin 2-\sin 1$

Vedclass · Answer

(A) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin (x-[x]) d x$. Since $[x]$ is the Greatest Integer Function,we split the integral based on the values of $[x]$ in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}] \approx [-1.57, 1.57]$.$I = \int_{-\frac{\pi}{2}}^{-1} \sin (x - (-2)) d x + \int_{-1}^{0} \sin (x - (-1)) d x + \int_{0}^{1} \sin (x - 0) d x + \int_{1}^{\frac{\pi}{2}} \sin (x - 1) d x$$I = \int_{-\frac{\pi}{2}}^{-1} \sin (x + 2) d x + \int_{-1}^{0} \sin (x + 1) d x + \int_{0}^{1} \sin x d x + \int_{1}^{\frac{\pi}{2}} \sin (x - 1) d x$$I = [-\cos (x + 2)]_{-\frac{\pi}{2}}^{-1} + [-\cos (x + 1)]_{-1}^{0} + [-\cos x]_{0}^{1} + [-\cos (x - 1)]_{1}^{\frac{\pi}{2}}$$I = -(\cos 1 - \cos(2 - \frac{\pi}{2})) - (\cos 1 - \cos 0) - (\cos 1 - \cos 0) - (\cos(\frac{\pi}{2} - 1) - \cos 0)$$I = -\cos 1 + \sin 2 - \cos 1 + 1 - \cos 1 + 1 - \sin 1 + 1$$I = 3 - 3\cos 1 + \sin 2 - \sin 1 = 3(1 - \cos 1) + \sin 2 - \sin 1$.

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin (x-[x]) d x=$,where $[.]$ denotes the Greatest Integer Function.

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