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Find the exact value of $$\int_{0}^{a} \frac{dx}{x + \sqrt{a^{2} - x^{2}}} $$ where $a$ is a positive constant.

The answer given to me is to use substitution. I have seen the answer, and feel that it is not intuitive. I would not have thought of it.

I was thinking of using some standard formulas to solve this problem, such as the following:

$$\int \frac{1}{a^{2} - x^{2}} dx = \frac{1}{2a}\ln\frac{a + x}{a - x} + C$$

$$\int \frac{1}{x^{2} - a^{2}} dx = \frac{1}{2a}\ln\frac{x - a}{x + a} + C$$

Is there any way to express the problem into these forms? I'm open to substitution as well.

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    Well, if you haven't observed, the standard *formulae* you are talking about, have all been derived using **substitution only**. – 2017-02-20
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    So you wish to avoid $x=a\sec(u)$? – 2017-02-20
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    @SchrodingersCat Hmm.. maybe I should edit my question. I'm open to substitution as well. – 2017-02-20
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    @SimplyBeautifulArt Yes, if it is possible. – 2017-02-20
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    @SimplyBeautifulArt The answer given to me was to use $x = asin\theta$. Your substitution is nicer. Could I say, as a general rule of thumb, that problems of these sorts, you use trigonometric substitution? – 2017-02-20
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    Yes, most definitely. – 2017-02-20
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    Actually, I made a mistake above. – 2017-02-20

2 Answers 2

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A standard substitution would be $x=a\sin t$, so $dx=a\cos t\,dt$ and $\sqrt{a^2-x^2}=a\cos t$ (assuming $a>0$). The integral thus becomes $$ I=\int_0^{\pi/2}\frac{\cos t}{\sin t+\cos t}\,dt $$ which admits a slick solution. Consider $$ \int_{0}^{\pi/2}\frac{\sin t}{\sin t+\cos t}\,dt= \Bigl[t=\frac{\pi}{2}-u\Bigr]= \int_{\pi/2}^{0}-\frac{\cos u}{\cos u+\sin u}\,du=I $$ Thus $$ 2I=\int_0^{\pi/2}\frac{\cos t}{\sin t+\cos t}\,dt+ \int_0^{\pi/2}\frac{\sin t}{\sin t+\cos t}\,dt= \int_{0}^{\pi/2}dt=\frac{\pi}{2} $$

If the integral had been with $\sqrt{x^2-a^2}$, then the logarithm would have appeared.

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    that was slick.....+1 – 2017-02-20
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    Haha, and this integral is on one of the current bounty questions! – 2017-02-20
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If we use $x=\frac a{\sqrt{1+t^2}}$, the integral reduces to

$$\int\frac1{x+\sqrt{a^2-x^2}}\ dx=-\int\frac t{(1+t)(1+t)^2}\ dt$$

And the rest is PFD.

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    Are you sure about this? I don't end up with the same expression as you after the substitution. – 2017-02-20
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    @mickep Sorry, I think I fixed it up now. – 2017-02-20
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    Since $0$t$. Perhaps you are more lucky with $x=a/\sqrt{1+t^2}$. – 2017-02-20
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    Yup, that sounds good. – 2017-02-20