# How can I find $\int_{\sqrt2/2}^{1}\int_{\sqrt{1-x^2}}^{x}\frac{1}{\sqrt{x^2+y^2}}dydx$?

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

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

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What have you tried so far? –  user17762 Jan 29 '11 at 15:17

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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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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Thanks for your alternative answer. –  MAxcoder Jan 29 '11 at 19:41
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. –  Andr Jan 30 '11 at 0:18

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

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Thank you very much. –  MAxcoder Jan 29 '11 at 19:22

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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