the integral from 0 to phi of sin^2 theta d(theta) =[theta/2 -sin(2theta)/4 ] from 0 to phi Can you explain to me how to arrive the second step ? I don't get it ! Thank you very much.
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I will use x instead of phi. $\displaystyle \int{sin^{2}(x)}dx$ Use the identity: $\displaystyle sin^{2}(x)=\frac{1}{2}(1-cos(2x))$ $\displaystyle \frac{1}{2}\int{[1-cos(2x)]}dx=\frac{1}{2}x-\frac{1}{4}sin(2x)+C$
galactus gave the simplest way, another one could be $\displaystyle \int\sin^2x~dx=\color{blue}\int\sin{x}\sin{x}~dx$ and apply integration by parts.
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