Indefinte integration reloaded

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21 Dec 2009 02:06:35 IST

Subject: Indefinte integration reloaded

akki ~~ unlucky forever ~~ (1635)

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$put\ x=sin2\theta\\ \\ I=\int ln|(|sin\theta-cos\theta)|+|sin\theta+cos\theta|)|cos2\theta\ d\theta\\ \\ as\ the\ interval\ is \ not\ specified,so\ two\ cases\ arises\\ \\ I=\int ln(2sin\theta)cos2\theta\ d\theta\ (when\ both\ mod\ opens\ with\ same\ sign)\\ \\ I=\int ln(2cos\theta)cos2\theta\ d\theta\ (when\ both \ mod\ opens\ with\ different\ sign)\\ \\ solving\ first\ case,integrating\ by-parts\\ \\ I=ln(2sin\theta).\frac{sin2\theta}{2}-\frac{(\theta+\frac{sin2\theta}{2})}{2}+C\\ \\ put\ back\ \theta\ in\ terms\ of\ x,get\ the\ ans\\ \\ solving\ second \ case,integrating\ by-parts\\ \\ I=ln(2sin\theta).\frac{sin2\theta}{2}+\frac{(\theta-\frac{sin2\theta}{2})}{2}+C\\ \\ put\ back\ \theta\ in\ terms\ of\ x,get\ the\ ans\\ \\$

all's well that end's well, but if it dosen't, then it is not the end . . .

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