- #1

planck42

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First step: Complexify the function in order to make straight integration possible

[tex]\sin x=\frac{e^{ix}-e^{-ix}}{2i}[/tex]

[tex]\sin^{2}x=-\frac{1}{2}-\frac{e^{2ix}+e^{-2ix}}{4}[/tex]Second step: Integrate the complex function

[tex]-\frac{1}{4}\int{2+e^{2ix}+e^{-2ix} dx} = -\frac{1}{4}(2x+\frac{e^{2ix}}{2i}-\frac{e^{-2ix}}{2i}+C)[/tex]

[tex]\mbox{However,} \frac{e^{2ix}-e^{-2ix}}{2i} = \sin(2x), \mbox{so}[/tex]

[tex]-\frac{1}{4}(2x+\frac{e^{2ix}}{2i}-\frac{e^{-2ix}}{2i}+C) = -\frac{\frac{1}{2}\sin(2x)+x}{2}+C, \mbox{which appears to be the answer. Can it survive the derivative test?}[/tex]Third step: Take the derivative of [tex]-\frac{\frac{1}{2}\sin(2x)+x}{2}+C[/tex] and see if it equals [tex]\sin^{2}x[/tex]

[tex]\frac{d}{dx}(-\frac{\frac{1}{2}\sin(2x)+x}{2}+C)=-\frac{1}{2}-\frac{1}{2}\cos(2x)=-\frac{1}{2}(1+\cos(2x))=-\cos^{2}x \mbox{, which is the answer less one.}[/tex]

Where is the error in the above process?