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Question Number 2970 by Syaka last updated on 01/Dec/15

$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\underset{\mathrm{0}} {\overset{{z}} {\int}}\underset{\mathrm{0}} {\overset{{y}} {\int}}\:{sin}\:\left({x}\:+\:{y}\:+\:{z}\right)\:{dx}\:{dy}\:{dz}\:=\:...? \\$$

Answered by Yozzi last updated on 02/Dec/15

$$\int{sin}\left({x}+{b}\right){dx}\:{let}\:{u}={x}+{b}\Rightarrow{du}={dx} \\$$$$\therefore\int{sinudu}=−{cosu}+{C}=−{cos}\left({x}+{b}\right)+{C} \\$$$$\therefore\:\int_{\mathrm{0}} ^{{y}} {sin}\left({x}+{y}+{z}\right){dx}=−{cos}\left({x}+{y}+{z}\right)\mid_{\mathrm{0}} ^{{y}} \\$$$$\int_{\mathrm{0}} ^{{y}} {sin}\left({x}+{y}+{z}\right){dx}={cos}\left({y}+{z}\right)−{cos}\left(\mathrm{2}{y}+{z}\right) \\$$$$\therefore\int_{\mathrm{0}} ^{{z}} \int_{\mathrm{0}} ^{{y}} {sin}\left({x}+{y}+{z}\right){dxdy} \\$$$$=\int_{\mathrm{0}} ^{{z}} {cos}\left({y}+{z}\right)−{cos}\left(\mathrm{2}{y}+{z}\right){dy} \\$$$$={sin}\left({y}+{z}\right)−\frac{\mathrm{1}}{\mathrm{2}}{sin}\left(\mathrm{2}{y}+{z}\right)\mid_{\mathrm{0}} ^{{z}} \\$$$$={sin}\mathrm{2}{z}−\mathrm{0}.\mathrm{5}{sin}\mathrm{3}{z}−\left({sinz}−\mathrm{0}.\mathrm{5}{sinz}\right) \\$$$$={sin}\mathrm{2}{z}−\mathrm{0}.\mathrm{5}{sin}\mathrm{3}{z}−\mathrm{0}.\mathrm{5}{sinz} \\$$$$\therefore\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \int_{\mathrm{0}} ^{{z}} \int_{\mathrm{0}} ^{{y}} {sin}\left({x}+{y}+{z}\right){dxdydz} \\$$$$=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {sin}\mathrm{2}{z}−\mathrm{0}.\mathrm{5}{sin}\mathrm{3}{z}−\mathrm{0}.\mathrm{5}{sinzdz} \\$$$$=−\mathrm{0}.\mathrm{5}{cos}\mathrm{2}{z}+\frac{\mathrm{1}}{\mathrm{6}}{cos}\mathrm{3}{z}+\mathrm{0}.\mathrm{5}{cosz}\mid_{\mathrm{0}} ^{\pi/\mathrm{2}} \\$$$$=−\mathrm{0}.\mathrm{5}\left(−\mathrm{1}\right)+\mathrm{0}+\mathrm{0}−\left(−\mathrm{0}.\mathrm{5}+\mathrm{0}.\mathrm{5}+\frac{\mathrm{1}}{\mathrm{6}}\right) \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{6}} \\$$$$=\frac{\mathrm{6}−\mathrm{2}}{\mathrm{12}} \\$$$$=\frac{\mathrm{1}}{\mathrm{3}} \\$$$$\\$$$$\\$$

Commented by Filup last updated on 02/Dec/15

$$\left.{damn},\:{you}\:{beat}\:{me}!\:;\right) \\$$

Commented by Syaka last updated on 02/Dec/15

$${Thanks}\:{Sir},\:{I}\:{always}\:{learn}\:{from}\:{you}. \\$$

Answered by Filup last updated on 03/Dec/15

$$\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\underset{\mathrm{0}} {\overset{{z}} {\int}}\underset{\mathrm{0}} {\overset{{y}} {\int}}\:\mathrm{sin}\left({x}+{y}+{z}\right)\:{dx}\:{dy}\:{dz} \\$$$$\\$$$$=\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\underset{\mathrm{0}} {\overset{{z}} {\int}}\:−\left[\mathrm{cos}\left({x}+{y}+{z}\right)\right]_{{x}=\mathrm{0}} ^{{x}={y}} \:{dy}\:{dz} \\$$$$=\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\underset{\mathrm{0}} {\overset{{z}} {\int}}\:\left(\mathrm{cos}\left({y}+{z}\right)−\mathrm{cos}\left(\mathrm{2}{y}+{z}\right)\right)\:{dy}\:{dz} \\$$$$=\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\:\left[\mathrm{sin}\left({y}+{z}\right)−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\left(\mathrm{2}{y}+{z}\right)\right]_{{y}=\mathrm{0}} ^{{y}={z}} \:{dz} \\$$$$=\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\:\left(\mathrm{sin}\left(\mathrm{2}{z}\right)−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\left(\mathrm{3}{z}\right)\right)−\left(\mathrm{sin}\left({z}\right)−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\left({z}\right)\right)\:{dz} \\$$$$=\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\:\left(\mathrm{sin}\left(\mathrm{2}{z}\right)−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\left(\mathrm{3}{z}\right)−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\left({z}\right)\right)\:{dz} \\$$$$=\left[−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\left(\mathrm{2}{z}\right)+\frac{\mathrm{1}}{\mathrm{6}}\mathrm{cos}\left(\mathrm{3}{z}\right)+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\left({z}\right)\right]_{{z}=\mathrm{0}} ^{{z}=\pi/\mathrm{2}} \\$$$$\\$$$$=\left(−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\left(\pi\right)+\frac{\mathrm{1}}{\mathrm{6}}\mathrm{cos}\left(\frac{\mathrm{3}\pi}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\left(\frac{\pi}{\mathrm{2}}\right)\right) \\$$$$−\left(−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\left(\mathrm{0}\right)+\frac{\mathrm{1}}{\mathrm{6}}\mathrm{cos}\left(\mathrm{0}\right)+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\left(\mathrm{0}\right)\right) \\$$$$\\$$$$\mathrm{cos}\left(\mathrm{0}\right)=\mathrm{1},\:\:\:\:\mathrm{cos}\left({n}\pi\right)=−\mathrm{1},\:\:\:\:\mathrm{cos}\left(\frac{{n}\pi}{\mathrm{2}}\right)=\mathrm{0},\:\:\:\:{n}\in\mathbb{Z} \\$$$$=\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{0}+\mathrm{0}\right)−\left(−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{2}}\right) \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{2}}{\mathrm{3}} \\$$$$=\mathrm{1}−\frac{\mathrm{2}}{\mathrm{3}} \\$$$$=\frac{\mathrm{1}}{\mathrm{3}} \\$$$$\\$$$$\therefore\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\underset{\mathrm{0}} {\overset{{z}} {\int}}\underset{\mathrm{0}} {\overset{{y}} {\int}}\:\mathrm{sin}\left({x}+{y}+{z}\right)\:{dx}\:{dy}\:{dz}\:=\:\frac{\mathrm{1}}{\mathrm{3}} \\$$

Commented by Syaka last updated on 02/Dec/15

$${Thanks}\:{for}\:{Solution}\:{Sir},\:{I}\:{like}\:{that} \\$$