solve : I = ∫_0 ^π (((r−R cosθ) sin θ )/((R^(2 ) + r^2 − 2Rr cos θ)^(3/2) )) dθ for r


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Question Number 33823 by 33 last updated on 25/Apr/18

$s o l v e :$ $I = \underset{0}{\int^{π}} \frac{(r - R c o s θ) s i n θ}{{(R^{2} + r^{2} - 2 R r c o s θ)}^{3 / 2}} d θ$ $f o r r < R$ $a n d r > R r e s p e c t i v e l y .$
	Answered by MJS last updated on 26/Apr/18

	$\int u^{'} v = u v - \int {u v}^{'}$ $u^{'} = \frac{\sin θ}{{(r^{2} + R^{2} - 2 r R \cos θ)}^{\frac{3}{2}}}; v = r - R \cos θ$ $u = - \frac{1}{r R {(r^{2} + R^{2} - 2 r R \cos θ)}^{\frac{1}{2}}}; v^{'} = R \sin θ$ $[\begin{matrix} \int \frac{\sin θ}{{(r^{2} + R^{2} - 2 r R \cos θ)}^{\frac{3}{2}}} d θ = \\ [t = r^{2} + R^{2} - 2 r R \cos θ \to d x = \frac{d t}{2 r R \sin θ}] \\ = \frac{1}{2 r R} \int \frac{1}{t^{\frac{3}{2}}} d t = - \frac{1}{{r R t}^{\frac{1}{2}}} \end{matrix}]$ $u v = - \frac{r - R \cos θ}{r R {(r^{2} + R^{2} - 2 r R \cos θ)}^{\frac{1}{2}}}$ $- \int {u v}^{'} = \int \frac{\sin θ}{r {(r^{2} + R^{2} - 2 r R \cos θ)}^{\frac{1}{2}}} =$ $[with the same substitution as above]$ $= \frac{{(r^{2} + R^{2} - 2 r R \cos θ)}^{\frac{1}{2}}}{r^{2} R}$ $\int \frac{(r - R \cos θ) \sin θ}{{(r^{2} + R^{2} - 2 r R \cos θ)}^{\frac{3}{2}}} =$ $= - \frac{r - R \cos θ}{r R {(r^{2} + R^{2} - 2 r R \cos θ)}^{\frac{1}{2}}} + \frac{{(r^{2} + R^{2} - 2 r R \cos θ)}^{\frac{1}{2}}}{r^{2} R} + C =$ $= \frac{R - r \cos θ}{r^{2} \sqrt{r^{2} + R^{2} - 2 r R \cos θ}} + C = F (θ)$ $F (π) - F (0) = \frac{R + r}{r^{2} ∣ R + r ∣} - \frac{R - r}{r^{2} ∣ R - r ∣}$ $r, R > 0$ $r < R \Rightarrow F (π) - F (0) = 0$ $r > R \Rightarrow F (π) - F (0) = \frac{2}{r^{2}}$
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