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I have a simple integral to evaluate. I am able to get the anti derivative using integration by substitution. However, the integral is a definite integral. Now, I have a question as to how to transform the limits of integration when a substitution is made:

The integral is: $$ \int_R^0 \sqrt {\dfrac{Rr} {R-r}} dr $$

I make a substitution $ r = R\sin^2A $

The integral then reduces to: $$ \int_a^b R^{3/2}\cdot(2\sin^2A)\ dA = R^{3/2}\Big[ A - \dfrac{\sin2A}{2} \Big]_a^b $$

(where $a$ and $b$ are the new limits).

Now, to calculate the new limits:

when $r = R\sin^2 A = 0$, then $ A = 0 $. Out of many possible values, I chose $ A = 0 $. When $r = R\sin^2 A = R$, then $ A = \pi/2 $ or $A = -pi/2$. Obviously the answer depends upon what values I chose for $A$. Now, here I am not choosing values of $A$ in which the periodic nature of the sine function comes into play. If we take the limits as $ \{ pi/2, 0 \} $ we get a different answer than the one we get by taking the limits as $ \{-pi/2, 0 \} $. Obviously something is wrong here. What is it?

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  • $\begingroup$ you have made a mistake. in your question $R$ appears as constant. in your substitution the same $R$ is taken as a variable . $\endgroup$ – Suraj M S Aug 28 '13 at 9:49
  • $\begingroup$ In my question, it isn't $R$ that is taken as a variable - it is a constant. Only $r$ and $A$ are the variables. $\endgroup$ – Parth Thakkar Aug 28 '13 at 17:53
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I believe the problem of choosing the limits is related to the simplification of the integrand. When you simplify $\sqrt{\frac{\sin^{2}A}{\cos^{2}A}}$ to get $\frac{\sin A}{\cos A}$, you are assuming that $\tan A=\frac{\sin A}{\cos A}\ge0$, so you have to choose your limits to make sure this is valid for your values of A.

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  • $\begingroup$ Hmm, that makes sense. I thought I would be able to find a counter example, but failed. +1! $\endgroup$ – Parth Thakkar Aug 29 '13 at 12:59
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in your substitution $A$ is the variable. $$r = R\sin^2(A)$$ $$dr = 2R\sin(A)\cos(A)dA $$ and further the limits changes to $ \frac{π}{2}$ to $0$ and the integral becomes further solvable.

you get the answer on solving as $$ I = -\frac{R^{\frac{3}{2}}π}{2}$$ this is what i get.

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  • $\begingroup$ My question is, why only $pi/2$ and $0$? Why can't I take other bounds which equally satisfy the relation $r = R\sin^2A$? Like $-pi/2$ and $0$? Different bounds give different answers. $\endgroup$ – Parth Thakkar Aug 28 '13 at 17:55

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