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My question is ; How can I solve the following integral question?

$\displaystyle \int_{\sqrt2/2}^{1}\int_{\sqrt{1-x^2}}^{x}\frac{1}{\sqrt{x^2+y^2}}dydx$

Thanks in advance,

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4 Answers 4

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You've had some time to study this, so let's look closer at the two evident approaches: (a) conversion to polar coordinates, (b) integrate directly with a standard hyperbolic substitution.

(a) Conversion to polar coordinates: Let

$$I = \int_{1/\sqrt{2}}^1 \, \int_{\sqrt{1-x^2}}^x \frac{1}{\sqrt{x^2+y^2}} \text{d}y \, \text{d}x.$$

If you draw a diagram of the situation you will see that

$$I = \int_0^{\pi/4} \, \int_{r=1}^{r=\sec \theta} \frac{1}{r} r \text{d}r \, \text{d} \theta = \int_0^{\pi/4} \sec \theta \, - 1 \textrm{ d} \theta $$

$$= \left[ \frac{1}{2} \log \left| \frac{1+ \sin \theta}{1- \sin \theta} \right| - \theta \right]_0^{\pi/4} = \log(1+\sqrt{2}) - \frac{\pi}{4}.$$

(b) Standard hyperbolic substitution: To evaluate the inner integral set $y= x \sinh u$ and noting that $\sinh^{-1} u = \log(u+ \sqrt{1+u^2})$ we have

$$\int_{\sqrt{1-x^2}}^x \frac{1}{\sqrt{x^2+y^2}} \text{d}y = \left[ \log \left( \frac{y}{x} + \sqrt{ 1+ \frac{y^2}{x^2} }\right) \right]_{\sqrt{1-x^2}}^x $$ $$= \log(1+ \sqrt{2}) + \log x - \log( 1+ \sqrt{1-x^2}). \quad (1)$$

Both the logs can be integrated by parts. The first is standard, $\int \log x \textrm{ d}x = x \log x - x + C$ and the second

$$ \int \log( 1+ \sqrt{1-x^2} ) \textrm{ d}x = x \log( 1+ \sqrt{1-x^2}) -x + \sin^{-1} x + C.$$

You will need (do the integration) to note that

$$\frac{1}{\sqrt{1-x^2}} - 1 = \frac{x^2}{(1-x^2) + \sqrt{1-x^2}}.$$

And so upon integrating $(1)$ between $1/\sqrt{2}$ and $1$ you obtain

$$I = \left(1 - \frac{1}{\sqrt{2}} \right) \log(1+ \sqrt{2}) + \left[ x \log x - x\log( 1+ \sqrt{1-x^2}) - \sin^{-1} x \right]_{1/\sqrt{2}}^1$$ $$= \log(1+\sqrt{2}) - \frac{\pi}{4}.$$

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Try converting to polar coordinates, you should end out integrating $\sec(\theta) - 1$.

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Hint:for the y integral, consider x to be a constant. It is a standard form that can be solved by a trigonometric substitution.

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Have you tried matlab? If you say:

int('1/sqrt(x^2+y^2)','y','sqrt(1-x^2)','x')

then you'll get: log(x + (2*x^2)^(1/2)) - log((1 - x^2)^(1/2) + 1) as first integral.

Then: int('log(x + (2*x^2)^(1/2)) - log((1 - x^2)^(1/2) + 1)', 'x')

Result:

x*(log(x + (2*x^2)^(1/2)) - 1) - int(log((1 - x^2)^(1/2) + 1), x)

But the last part can't be integrated symbolically, so, in matlab, you can evaluate it numerically:

`double(int('log((1 - x^2)^(1/2) + 1)', 'x',sqrt(2)/2,1))` 

which will give : ans = 0.1143, and the other part you can evaluate yourself. It should be easy to replace x with the 2 numbers. Hope this helps!

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  • $\begingroup$ Thanks for your alternative answer. $\endgroup$
    – MAxcoder
    Jan 29, 2011 at 19:41
  • $\begingroup$ Have you checked it? Is it ok? I haven't checked it, i just copied from matlab. It's quite cool. You can also try wolframalpha, it shows the steps, but it didn't work for me (computation timeout), heh, it was to tough for it. $\endgroup$
    – Andr
    Jan 30, 2011 at 0:18

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