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baha jaff Grade: 12
         int(from (0) to (pi) Ln[(b-cosx) / (a-cosx)]dx         a,b>0
6 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 & Regards
Jitender Singh
IIT Delhi
askIITians Faculty
3 years ago
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