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Question Number 27280 by kemhoney78@gmail.com last updated on 04/Jan/18

$$\mathrm{If}\mathrm{for}\mathrm{a}\mathrm{real}\mathrm{number}y,\left[y\right]\mathrm{is}\mathrm{the}\mathrm{greatest}$$$$\mathrm{integer}\mathrm{less}\mathrm{than}\mathrm{or}\mathrm{equal}\mathrm{to}y,\mathrm{then}\mathrm{the}$$$$\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{integeral}\underset{\pi /2}{\stackrel{3\pi /2}{\int }}\left[2\sin x\right]dx\mathrm{is}$$

Commented by abdo imad last updated on 04/Jan/18

$$letdothechangementx=\frac{\pi }{2}+t$$$${\int }_{\frac{\pi }{2}}^{\frac{3\pi }{2}}\left[2sinx\right]dx={\int }_0^{\pi }\left[2cosx\right]dx(2cosx=t\iff x=arcos\left(\frac{t}{2}\right))=$$$$={\int }_2^{-2}\left[t\right](-\frac{1}{2\sqrt{1-\frac{t^2}{4}}})dt=\frac{1}{2}{\int }_{-2}^2\frac{\left[t\right]}{\sqrt{1-\frac{t^2}{4}}}dt$$$$=\frac{1}{2}{\int }_{-2}^{-1}(...)dt+\frac{1}{2}{\int }_{-1}^0(...)dt+\frac{1}{2}{\int }_0^1(...)dt+\frac{1}{2}{\int }_1^2(...)dt$$$$=-{\int }_{-2}^{-1}\frac{dt}{\sqrt{1-\frac{t^2}{4}}}-\frac{1}{2}{\int }_{-1}^0\frac{dt}{\sqrt{1-\frac{t^2}{4}}}+0+\frac{1}{2}{\int }_1^2\frac{dt}{\sqrt{1-\frac{t^2}{4}}}dt$$$$=-{\int }_2^1\frac{-dt}{\sqrt{1-\frac{t^2}{4}}}+\frac{1}{2}{\int }_1^2\frac{dt}{\sqrt{1-\frac{t^2}{4}}}-\frac{1}{2}{\int }_1^0\frac{-dt}{\sqrt{1-\frac{t^2}{4}}}$$$$=-{\int }_1^2\frac{dt}{\sqrt{1-\frac{t^2}{4}}}+\frac{1}{2}{\int }_1^2\frac{dt}{\sqrt{1-\frac{t^2}{2}}}-\frac{1}{2}{\int }_0^1\frac{dt}{\sqrt{1-\frac{t^2}{4}}}$$$$=-\frac{1}{2}{\int }_1^2\frac{dt}{\sqrt{1-\frac{t^2}{2}}}-\frac{1}{2}{\int }_0^1\frac{dt}{\sqrt{1-\frac{t^2}{4}}}dt$$$$=-\frac{1}{2}{\int }_0^2\frac{dt}{\sqrt{1-\frac{t^2}{4}}}dt={\left[arcos\left(\frac{t}{2}\right)\right]}_0^2$$$$=arcos\left(1\right)-arcos0=-\frac{\pi }{2}$$

Answered by ajfour last updated on 04/Jan/18

$$I={\int }_{\pi /2}^{5\pi /6}dx+{\int }_{5\pi /6}^{\pi }\left(0\right)dx+{\int }_{\pi }^{7\pi /6}(-1)dx$$$$+{\int }_{7\pi /6}^{3\pi /2}(-2)dx$$$$=\frac{\pi }{3}+0-\frac{\pi }{6}-\frac{2\pi }{3}=-\frac{\pi }{2}.$$

Commented by ajfour last updated on 04/Jan/18

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