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

$$\underset{0}{\stackrel{\frac{\pi }{2}}{\int }}\underset{0}{\stackrel{z}{\int }}\underset{0}{\stackrel{y}{\int }}sin(x+y+z)dxdydz=...?$$

Answered by Yozzi last updated on 02/Dec/15

$$\int sin(x+b)dxletu=x+b\Rightarrow du=dx$$$$\therefore \int sinudu=-cosu+C=-cos(x+b)+C$$$$\therefore {\int }_0^{y}sin(x+y+z)dx=-cos(x+y+z){\mid }_0^y$$$${\int }_0^{y}sin(x+y+z)dx=cos(y+z)-cos(2y+z)$$$$\therefore {\int }_0^z{\int }_0^{y}sin(x+y+z)dxdy$$$$={\int }_0^{z}cos(y+z)-cos(2y+z)dy$$$$=sin(y+z)-\frac{1}{2}sin(2y+z){\mid }_0^z$$$$=sin2z-0.5sin3z-(sinz-0.5sinz)$$$$=sin2z-0.5sin3z-0.5sinz$$$$\therefore {\int }_0^{\pi /2}{\int }_0^z{\int }_0^{y}sin(x+y+z)dxdydz$$$$={\int }_0^{\pi /2}sin2z-0.5sin3z-0.5sinzdz$$$$=-0.5cos2z+\frac{1}{6}cos3z+0.5cosz{\mid }_0^{\pi /2}$$$$=-0.5(-1)+0+0-(-0.5+0.5+\frac{1}{6})$$$$=\frac{1}{2}-\frac{1}{6}$$$$=\frac{6-2}{12}$$$$=\frac{1}{3}$$$$$$$$$$

Commented by Filup last updated on 02/Dec/15

$$damn,youbeatme!;)$$

Commented by Syaka last updated on 02/Dec/15

$$ThanksSir,Ialwayslearnfromyou.$$

Answered by Filup last updated on 03/Dec/15

$$\underset{0}{\stackrel{\pi /2}{\int }}\underset{0}{\stackrel{z}{\int }}\underset{0}{\stackrel{y}{\int }}\sin (x+y+z)dxdydz$$$$$$$$=\underset{0}{\stackrel{\pi /2}{\int }}\underset{0}{\stackrel{z}{\int }}-{\left[\cos (x+y+z)\right]}_{x=0}^{x=y}dydz$$$$=\underset{0}{\stackrel{\pi /2}{\int }}\underset{0}{\stackrel{z}{\int }}(\cos (y+z)-\cos (2y+z))dydz$$$$=\underset{0}{\stackrel{\pi /2}{\int }}{[\sin (y+z)-\frac{1}{2}\sin (2y+z)]}_{y=0}^{y=z}dz$$$$=\underset{0}{\stackrel{\pi /2}{\int }}(\sin \left(2z\right)-\frac{1}{2}\sin \left(3z\right))-(\sin \left(z\right)-\frac{1}{2}\sin \left(z\right))dz$$$$=\underset{0}{\stackrel{\pi /2}{\int }}(\sin \left(2z\right)-\frac{1}{2}\sin \left(3z\right)-\frac{1}{2}\sin \left(z\right))dz$$$$={[-\frac{1}{2}\cos \left(2z\right)+\frac{1}{6}\cos \left(3z\right)+\frac{1}{2}\cos \left(z\right)]}_{z=0}^{z=\pi /2}$$$$$$$$=(-\frac{1}{2}\cos \left(\pi \right)+\frac{1}{6}\cos \left(\frac{3\pi }{2}\right)+\frac{1}{2}\cos \left(\frac{\pi }{2}\right))$$$$-(-\frac{1}{2}\cos \left(0\right)+\frac{1}{6}\cos \left(0\right)+\frac{1}{2}\cos \left(0\right))$$$$$$$$\cos \left(0\right)=1,\cos \left(n\pi \right)=-1,\cos \left(\frac{n\pi }{2}\right)=0,n\in \mathbb{Z}$$$$=(\frac{1}{2}+0+0)-(-\frac{1}{2}+\frac{1}{6}+\frac{1}{2})$$$$=\frac{1}{2}+\frac{1}{2}-\frac{2}{3}$$$$=1-\frac{2}{3}$$$$=\frac{1}{3}$$$$$$$$\therefore \underset{0}{\stackrel{\pi /2}{\int }}\underset{0}{\stackrel{z}{\int }}\underset{0}{\stackrel{y}{\int }}\sin (x+y+z)dxdydz=\frac{1}{3}$$

Commented by Syaka last updated on 02/Dec/15

$$ThanksforSolutionSir,Ilikethat$$

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