r = 1 + sin(theta) i know the integral equation should be: 1/2 integral pi/6 -> pi/2 ( 1 + sin(theta) + sin^2(theta) ) just need help solving through the integral please. thanks.
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Hello rcmango Originally Posted by rcmango r = 1 + sin(theta) i know the integral equation should be: 1/2 integral pi/6 -> pi/2 ( 1 + sin(theta) + sin^2(theta) ) just need help solving through the integral please. thanks. To integrate the $\displaystyle \sin^2\theta$ term, use $\displaystyle \cos 2\theta = 1 - 2\sin^2\theta$ in the form $\displaystyle \sin^2\theta = \tfrac{1}{2}(1 - \cos 2\theta)$ and it's very straightforward from there. Grandad
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