Question Number 27280 by kemhoney78@gmail.com last updated on 04/Jan/18
![If for a real number y, [y] is the greatest integer less than or equal to y, then the value of the integeral ∫_(π/2) ^(3π/2) [2 sin x]dx is](https://www.tinkutara.com/question/Q27280.png)
$$\mathrm{If}\:\mathrm{for}\:\mathrm{a}\:\mathrm{real}\:\mathrm{number}\:{y},\:\left[{y}\right]\:\mathrm{is}\:\mathrm{the}\:\mathrm{greatest} \\ $$$$\mathrm{integer}\:\mathrm{less}\:\mathrm{than}\:\mathrm{or}\:\mathrm{equal}\:\mathrm{to}\:{y},\:\mathrm{then}\:\mathrm{the} \\ $$$$\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{integeral}\:\underset{\pi/\mathrm{2}} {\overset{\mathrm{3}\pi/\mathrm{2}} {\int}}\left[\mathrm{2}\:\mathrm{sin}\:{x}\right]{dx}\:\mathrm{is} \\ $$
Commented by abdo imad last updated on 04/Jan/18
![let do the changement x= (π/2) +t ∫_(π/2) ^((3π)/2) [2sinx]dx= ∫_0 ^π [2 cosx]dx ( 2cosx=t ⇔ x =arcos((t/2)))= =∫_2 ^(−2) [t](−(1/(2(√(1−(t^2 /4))))))dt= (1/2) ∫_(−2) ^2 (([t])/( (√(1−(t^2 /4))))) dt =(1/2) ∫_(−2) ^(−1) (...)dt +(1/2) ∫_(−1) ^0 (...)dt +(1/2) ∫_0 ^1 (...)dt +(1/2) ∫_1 ^2 (...)dt =−∫_(−2) ^(−1 ) (dt/( (√(1−(t^2 /4))))) −(1/2) ∫_(−1) ^0 (dt/( (√(1−(t^2 /4))))) +0+(1/2)∫_1 ^2 (dt/( (√(1−(t^2 /4)))))dt =−∫_2 ^1 ((−dt)/( (√(1−(t^2 /4))))) +(1/2)∫_1 ^(2 ) (dt/( (√(1−(t^2 /4))))) −(1/2) ∫_1 ^0 ((−dt)/( (√(1−(t^2 /4))))) =−∫_1 ^2 (dt/( (√(1−(t^2 /4))))) +(1/2) ∫_1 ^2 (dt/( (√(1−(t^2 /2))))) −(1/2) ∫_0 ^1 (dt/( (√(1−(t^2 /4))))) =−(1/2) ∫_1 ^2 (dt/( (√(1−(t^2 /2))))) −(1/2) ∫_0 ^1 (dt/( (√(1−(t^2 /4)))))dt =−(1/2) ∫_0 ^2 (dt/( (√(1−(t^2 /4))))) dt=[ arcos((t/2))]_0 ^2 =arcos(1)−arcos0= −(π/2)](https://www.tinkutara.com/question/Q27292.png)
$${let}\:{do}\:{the}\:{changement}\:{x}=\:\frac{\pi}{\mathrm{2}}\:+{t} \\ $$$$\int_{\frac{\pi}{\mathrm{2}}} ^{\frac{\mathrm{3}\pi}{\mathrm{2}}} \left[\mathrm{2}{sinx}\right]{dx}=\:\int_{\mathrm{0}} ^{\pi} \left[\mathrm{2}\:{cosx}\right]{dx}\:\left(\:\mathrm{2}{cosx}={t}\:\Leftrightarrow\:{x}\:={arcos}\left(\frac{{t}}{\mathrm{2}}\right)\right)= \\ $$$$=\int_{\mathrm{2}} ^{−\mathrm{2}} \left[{t}\right]\left(−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{1}−\frac{{t}^{\mathrm{2}} }{\mathrm{4}}}}\right){dt}=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{−\mathrm{2}} ^{\mathrm{2}} \:\:\:\:\:\frac{\left[{t}\right]}{\:\sqrt{\mathrm{1}−\frac{{t}^{\mathrm{2}} }{\mathrm{4}}}}\:{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{−\mathrm{2}} ^{−\mathrm{1}} \left(…\right){dt}\:+\frac{\mathrm{1}}{\mathrm{2}}\:\int_{−\mathrm{1}} ^{\mathrm{0}} \left(…\right){dt}\:\:+\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(…\right){dt}\:+\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{1}} ^{\mathrm{2}} \left(…\right){dt} \\ $$$$=−\int_{−\mathrm{2}} ^{−\mathrm{1}\:} \:\frac{{dt}}{\:\sqrt{\mathrm{1}−\frac{{t}^{\mathrm{2}} }{\mathrm{4}}}}\:−\frac{\mathrm{1}}{\mathrm{2}}\:\int_{−\mathrm{1}} ^{\mathrm{0}} \:\frac{{dt}}{\:\sqrt{\mathrm{1}−\frac{{t}^{\mathrm{2}} }{\mathrm{4}}}}\:+\mathrm{0}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{1}} ^{\mathrm{2}} \frac{{dt}}{\:\sqrt{\mathrm{1}−\frac{{t}^{\mathrm{2}} }{\mathrm{4}}}}{dt} \\ $$$$=−\int_{\mathrm{2}} ^{\mathrm{1}} \frac{−{dt}}{\:\sqrt{\mathrm{1}−\frac{{t}^{\mathrm{2}} }{\mathrm{4}}}}\:+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{1}} ^{\mathrm{2}\:} \frac{{dt}}{\:\sqrt{\mathrm{1}−\frac{{t}^{\mathrm{2}} }{\mathrm{4}}}}\:−\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{1}} ^{\mathrm{0}} \frac{−{dt}}{\:\sqrt{\mathrm{1}−\frac{{t}^{\mathrm{2}} }{\mathrm{4}}}} \\ $$$$=−\int_{\mathrm{1}} ^{\mathrm{2}} \frac{{dt}}{\:\sqrt{\mathrm{1}−\frac{{t}^{\mathrm{2}} }{\mathrm{4}}}}\:\:+\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{1}} ^{\mathrm{2}} \frac{{dt}}{\:\sqrt{\mathrm{1}−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}}}\:−\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dt}}{\:\sqrt{\mathrm{1}−\frac{{t}^{\mathrm{2}} }{\mathrm{4}}}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{1}} ^{\mathrm{2}} \:\frac{{dt}}{\:\sqrt{\mathrm{1}−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}}}\:−\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dt}}{\:\sqrt{\mathrm{1}−\frac{{t}^{\mathrm{2}} }{\mathrm{4}}}}{dt} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{2}} \frac{{dt}}{\:\sqrt{\mathrm{1}−\frac{{t}^{\mathrm{2}} }{\mathrm{4}}}}\:{dt}=\left[\:{arcos}\left(\frac{{t}}{\mathrm{2}}\right)\right]_{\mathrm{0}} ^{\mathrm{2}} \\ $$$$={arcos}\left(\mathrm{1}\right)−{arcos}\mathrm{0}=\:−\frac{\pi}{\mathrm{2}} \\ $$
Answered by ajfour last updated on 04/Jan/18
$${I}=\int_{\pi/\mathrm{2}} ^{\:\:\mathrm{5}\pi/\mathrm{6}} {dx}+\int_{\mathrm{5}\pi/\mathrm{6}} ^{\:\:\pi} \left(\mathrm{0}\right){dx}+\int_{\pi} ^{\:\:\mathrm{7}\pi/\mathrm{6}} \left(−\mathrm{1}\right){dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:+\int_{\mathrm{7}\pi/\mathrm{6}} ^{\:\:\mathrm{3}\pi/\mathrm{2}} \left(−\mathrm{2}\right){dx} \\ $$$$\:=\frac{\pi}{\mathrm{3}}+\mathrm{0}−\frac{\pi}{\mathrm{6}}−\frac{\mathrm{2}\pi}{\mathrm{3}}\:=−\frac{\pi}{\mathrm{2}}\:. \\ $$
Commented by ajfour last updated on 04/Jan/18
