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question mark

int(from (0) to (pi) Ln[(b-cosx) / (a-cosx)]dxa,b>0

baha jaff , 14 Years ago

Grade 12

1 Answers

Jitender Singh

Last Activity: 10 Years ago

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

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