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I was trying to prove that the electric field at a distance $x>R$ from the centre of a charged shell is $\dfrac{kQ}{x^2}$. I did the physics part but I got stuck when I encountered the following integral:($R$ and $x$ are constants) $$\frac{kQ}{2}\int_0^\pi{(x-R\cos\theta)\sin\theta \over (x^2+R^2-2Rx\cos\theta)^{1.5}}d\theta$$ I only know the very basic integration techniques, like trigonometric substituion, or by parts integration so it would be much helpful if someone can give a hint on how to start with that.

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  • $\begingroup$ A little background on how you got this integral would be nice. $\endgroup$ – Arpan Apr 9 '15 at 5:55
  • $\begingroup$ @Arpan the background is irrelevant to the evaluation of this integral, but I'll post it if more people want to know about it. $\endgroup$ – G-man Apr 9 '15 at 7:42
  • $\begingroup$ I know, Just said it would be nice to see what method you were using. $\endgroup$ – Arpan Apr 9 '15 at 7:43
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    $\begingroup$ Use in some way my answer. $\endgroup$ – Tony Piccolo Apr 9 '15 at 11:12
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Okay I figured it out with Tony Piccolo's hint. Just so I have closure, I am answering my question.

Employ the substitution $u^2=(x^2+R^2-2Rx\cos\theta)\implies u\dfrac{du}{d\theta}=Rx\sin\theta$

Now our integral can become: $$\frac{kQ}{2}\int{(x-R\cos\theta)\sin\theta \over u^3}\times{udu \over Rx\sin\theta}=\frac{kQ}{2Rx}\int {(x-R\cos\theta) \over u^2}du$$ $$=\frac{kQ}{2Rx}\int \left({x-\frac{x^2+R^2-u^2}{2x} \over u^2}\right)du=\frac{kQ}{4Rx^2}\int {(u^2+x^2-R^2) \over u^2}du$$ $$=\frac{kQ}{4Rx^2}\left( \int du+(x^2-R^2)\int \frac{du}{u^2} \right)=\frac{kQ}{4Rx^2}\left( {u^2-x^2+R^2\over u} \right)$$

Now applying the limits $(x-R)\to(x+R)$ on the the result, we get $$\frac{kQ}{4Rx^2}\left( {2R^2+2Rx\over R+x}-{2R^2-2Rx\over x-R} \right)={kQ\over x^2}$$

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