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The textbook says that

$$\int\frac{-mu}{\sqrt{1-\left(\frac{u}{c}\right)^{2}}}du$$
can be solved by inspection to give $$mc^{2}\sqrt{1-\left(\frac{u}{c}\right)^{2}}$$

In simple steps, could anyone please explain how this is done.

Thank you

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4  
The integrand has the form $\frac{f'}{\sqrt f}$ (up to a multiplicative constant), so it's the derivative of $\sqrt f$. –  Davide Giraudo Sep 28 '11 at 11:13
1  
At least by inspection one can see that the answer is right! To proceed in the other direction, one notices that the derivative of $1-(u/c)^2$ is (basically) sitting upstairs, so the answer will be, more or less, $\sqrt{1-(u/c)^2}$. Then I would mentally differentiate to decide on the constant. –  André Nicolas Sep 28 '11 at 14:21
2  
Well, integration by inspection proceeds via the following simple steps: (1) Inspect. (2) Integrate. :-) –  Hans Lundmark Sep 28 '11 at 17:51

1 Answer 1

up vote 5 down vote accepted

You can always check whether Wolfram|Alpha shows you the steps for an integration; it does in this case (click "Show Steps").

Note that the integrand is proportional to $f'(g(u))g'(u)$ with $f(x)=\sqrt x$ and $g(u)=1-\left(\frac uc\right)^2$. Generally, if an integrand is a product of some expression and the derivative of some part of that expression, try substituting for that part.

I wouldn't worry too much about the fact that it says "by inspection" and it doesn't feel like mere inspection to you. If you do enough substitutions, over time you'll start to "see" this immediately and will be able to write down the integral without going through the mechanics of the substitution.

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Excellent answer. I never knew Wolfram|Alpha even existed, but it's brilliant, clearly showing all the steps. Thank you –  Peter4075 Sep 28 '11 at 13:15

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