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baha jaff Grade: 12
        int(from  (0) to (pi)
Ln[(b-cosx) / (a-cosx)] dx
7 years ago

## Answers : (1)

Jitender Singh
IIT Delhi
askIITians Faculty
158 Points
										Ans:$I = \int_{0}^{\pi}ln(\frac{b-cosx}{a-cosx})dx$…..............(1)$I = \int_{0}^{\pi}ln(\frac{b-cos(\pi -x)}{a-cos(\pi -x)})dx$$I = \int_{0}^{\pi}ln(\frac{b+cos(x)}{a+cos(x)})dx$.............(2)(1) + (2)$2I = \int_{0}^{\pi}ln(\frac{b^{2}-cos^{2}(x)}{a^{2}-cos^{2}(x)})dx$Integration by Parts$2I = (x.ln(\frac{b^{2}-cos^{2}(x)}{a^{2}-cos^{2}(x)}))_{0}^{\pi }-\int_{0}^{\pi}\frac{x.(a^{2}-cos^{2}x)}{b^{2}-cos^{2}x}dx$$I = -\frac{1}{2}\int_{0}^{\pi}\frac{x.(a^{2}-cos^{2}x)}{b^{2}-cos^{2}x}dx$$I = -\frac{1}{2}\int_{0}^{\pi}\frac{(\pi -x).(a^{2}-cos^{2}(\pi -x))}{b^{2}-cos^{2}(\pi -x)}dx$$I = -\frac{1}{2}\int_{0}^{\pi}\frac{(\pi -x).(a^{2}-cos^{2}(x))}{b^{2}-cos^{2}(x)}dx$$2I = -\frac{\pi}{2}\int_{0}^{\pi}\frac{(a^{2}-cos^{2}(x))}{b^{2}-cos^{2}(x)}dx$$I = -\frac{\pi}{4}\int_{0}^{\pi}\frac{(a^{2}-cos^{2}(x))}{b^{2}-cos^{2}(x)}dx$$I = -\frac{\pi}{4}\int_{0}^{\pi}\frac{(a^{2}-cos^{2}(x))}{b^{2}-cos^{2}(x)}.\frac{sec^{4}x}{sec^{4}x}dx$$I = -\frac{\pi}{4}\int_{0}^{\pi}\frac{(a^{2}sec^{4}x-sec^{2}x)}{b^{2}sec^{4}x-sec^{2}x}dx$$t = tanx$$dt = sec^{2}x.dx$$I = -\frac{\pi}{4}\int\frac{(a^{2}(t^{2}+1)-1)}{(t^{2}+1).(b^{2}(t^{2}+1)-1)}dt$Simply using the partial fraction rule here, we have$I = -\frac{\pi}{4}.(\frac{(a^{2}-b^{2})tan^{-1}(\frac{bt}{\sqrt{b^{2}-1}})+b\sqrt{b^{2}-1}tan^{-1}t}{b\sqrt{b^{2}-1}})$$I = -\frac{\pi}{4}.(\frac{(a^{2}-b^{2})tan^{-1}(\frac{btanx}{\sqrt{b^{2}-1}})+b\sqrt{b^{2}-1}tan^{-1}tanx}{b\sqrt{b^{2}-1}})$$I = -\frac{\pi}{4}.(\frac{(a^{2}-b^{2})tan^{-1}(\frac{btanx}{\sqrt{b^{2}-1}})}{b\sqrt{b^{2}-1}})+x)_{0}^{\pi}$$I = -\frac{\pi }{4}.\pi$$I = -\frac{\pi^{2} }{4}$Thanks & RegardsJitender SinghIIT DelhiaskIITians Faculty

3 years ago
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