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I've encountered the following problem in Morris Kline's textbook on Calculus (chapter 10, section 5, ex. 2) that I can't seem to understand-

if y' = sin x cos x, then if I set u = sin x, then du/dx = cos x (chain rule), then y = (sin^2 x) / 2. If I set u = cos x, then du/dx= -sin x, so I multiply the RHS of y' by -1/-1 and eventually arrive at y = -(cos^2 x) / 2. Then y = (sin^2 x) / 2 = - (cos^2 x) / 2. What's wrong?

I've been getting other problems in this section, but this one has me stumped . . . . plus I feel that I'm missing something important. Thanks, EW

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# Trig integration problem

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