the integral (between pi/6 and 0) of the integral (between pi/3 and 0) xsin(x+y)dydx
L lou_skywalker New member Joined Oct 30, 2009 Messages 3 Oct 30, 2009 #1 the integral (between pi/6 and 0) of the integral (between pi/3 and 0) xsin(x+y)dydx
B BigGlenntheHeavy Senior Member Joined Mar 8, 2009 Messages 1,577 Oct 31, 2009 #2 ∫0π/6∫0π/3xsin(x+y)dydx = ∫0π/6−xcos(x+y)]0π/3dx\displaystyle \int_ {0}^{\pi/6}\int_ {0}^{\pi/3}xsin(x+y)dydx \ = \ \int_{0}^{\pi/6}-xcos(x+y)\bigg]_{0}^{\pi/3}dx∫0π/6∫0π/3xsin(x+y)dydx = ∫0π/6−xcos(x+y)]0π/3dx = ∫0π/6[xcos(x)−xcos(x+π/3)]dx = 12∫0π/6xcos(x)dx+32∫0π/6xsin(x)dx\displaystyle = \ \int_{0}^{\pi/6}[xcos(x)-xcos(x+\pi/3)]dx \ = \ \frac{1}{2}\int_{0}^{\pi/6}xcos(x)dx+\frac{\sqrt3}{2}\int_{0}^{\pi/6}xsin(x)dx= ∫0π/6[xcos(x)−xcos(x+π/3)]dx = 21∫0π/6xcos(x)dx+23∫0π/6xsin(x)dx = .063912395792+.040313620193 = .104226015985\displaystyle = \ .063912395792+.040313620193 \ = \ .104226015985= .063912395792+.040313620193 = .104226015985 Note: In the second row, expand cos(x+π/3) and then use integration by parts.\displaystyle Note: \ In \ the \ second \ row, \ expand \ cos(x+\pi/3) \ and \ then \ use \ integration \ by \ parts.Note: In the second row, expand cos(x+π/3) and then use integration by parts. Afterthought: Or, if you have a trusty TI−89, just plug in the double integral and, walla,\displaystyle Afterthought: \ Or, \ if \ you \ have \ a \ trusty \ TI-89, \ just \ plug \ in \ the \ double \ integral \ and, \ walla,Afterthought: Or, if you have a trusty TI−89, just plug in the double integral and, walla, you have your answer.\displaystyle you \ have \ your \ answer.you have your answer.
∫0π/6∫0π/3xsin(x+y)dydx = ∫0π/6−xcos(x+y)]0π/3dx\displaystyle \int_ {0}^{\pi/6}\int_ {0}^{\pi/3}xsin(x+y)dydx \ = \ \int_{0}^{\pi/6}-xcos(x+y)\bigg]_{0}^{\pi/3}dx∫0π/6∫0π/3xsin(x+y)dydx = ∫0π/6−xcos(x+y)]0π/3dx = ∫0π/6[xcos(x)−xcos(x+π/3)]dx = 12∫0π/6xcos(x)dx+32∫0π/6xsin(x)dx\displaystyle = \ \int_{0}^{\pi/6}[xcos(x)-xcos(x+\pi/3)]dx \ = \ \frac{1}{2}\int_{0}^{\pi/6}xcos(x)dx+\frac{\sqrt3}{2}\int_{0}^{\pi/6}xsin(x)dx= ∫0π/6[xcos(x)−xcos(x+π/3)]dx = 21∫0π/6xcos(x)dx+23∫0π/6xsin(x)dx = .063912395792+.040313620193 = .104226015985\displaystyle = \ .063912395792+.040313620193 \ = \ .104226015985= .063912395792+.040313620193 = .104226015985 Note: In the second row, expand cos(x+π/3) and then use integration by parts.\displaystyle Note: \ In \ the \ second \ row, \ expand \ cos(x+\pi/3) \ and \ then \ use \ integration \ by \ parts.Note: In the second row, expand cos(x+π/3) and then use integration by parts. Afterthought: Or, if you have a trusty TI−89, just plug in the double integral and, walla,\displaystyle Afterthought: \ Or, \ if \ you \ have \ a \ trusty \ TI-89, \ just \ plug \ in \ the \ double \ integral \ and, \ walla,Afterthought: Or, if you have a trusty TI−89, just plug in the double integral and, walla, you have your answer.\displaystyle you \ have \ your \ answer.you have your answer.
