Calculate the Double Integral

[jenn9580]

I began integrating in respect to y, so that 5x could be pulled out as a constant. Then I used substitution. Then I integrated in respect to x. My final answer was wrong according to my online homework. I believe my problem is a simple mistake in integration but I am not seeing it.

 

[DrMike]

Nor can we, at the moment...

... but we might be able to help if you could type in some of your intermediate steps....

 

[BigGlenntheHeavy]

\(\displaystyle \int\int_{R}5xsin(x+y)dA \ \ R \ = \ \bigg[0,\frac{\pi}{6}\bigg] \ X \ \bigg[0,\frac{\pi}{3}\bigg]\)

\(\displaystyle = \ \int_{0}^{\pi/6}\int_{0}^{\pi/3}5xsin(x+y)dydx \ = \ \int_{0}^{\pi/6}-5xcos(x+y)\bigg]_{0}^{\pi/3}dx\)

\(\displaystyle = \ \int_{0}^{\pi/6}[5xcos(x)-5xcos(x+\pi/3)]dx \ = \ \frac{5}{2}\int_{0}^{\pi/6}xcos(x)dx+\frac{5\sqrt3}{2}\int_{0}^{\pi/6}xsin(x)dx\)

\(\displaystyle = \ .521130079926\)

\(\displaystyle Note: \ In \ the \ third \ row, \ expand \ cos(x+\pi/3) \ and \ then \ use \ integration \ by \ parts.\)

 

[jenn9580]

Thanks!! I made a rookie mistake! I didn't expand the first cos(x+y) correctly on line 2. I feel like an idiot.

 

[jenn9580]

Nevermind. I see it now!