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Find the exact value of

$$ \int_{0}^{a} \frac{1}{\sqrt {a^2-x^2}} \mathrm {dx} $$

Where, $a$ is a positive constant

Hi, guys can give me tips to solve this ? Should we use like u substitution?

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  • $\begingroup$ yes use u-subs. look at trig/hyperbolic functions to equations where you have quadratic forms of the following form $$a^2+x^2\\a^2-x^2$$ etc.. $\endgroup$
    – Chinny84
    Commented Feb 8, 2016 at 11:38

3 Answers 3

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Here is a tip,

for $a>0$

Factor a out: $$\frac {1} {a}\int_{0}^{a}\frac{1}{\sqrt{1-\dfrac{x^2}{a}}}\, \mathrm dx$$

And then set $t=x/a$ threfore $\mathrm dt=\frac 1 a\,\mathrm dx$

$$\int_{0}^{1}\frac{1}{\sqrt{1-t^2}}\,\mathrm dt\\ =\arcsin\left(\frac x a\right)\bigg|_0 ^a\\ $$


for $a<0$

switch the order of the integration bounds, factor $|a|$ out:

$$-\frac {1} {|a|}\int_{-a}^{0}\frac{1}{\sqrt{1-\dfrac{x^2}{a}}}\, \mathrm dx$$

And then set $t=x/a$ threfore $\mathrm dt=\frac 1 a\,\mathrm dx$

$$=-\int_{-1}^{0}\frac{1}{\sqrt{1-t^2}}\,\mathrm dt\\ =(-\arcsin\left(\frac x a\right))\bigg|_{-a} ^{0}\\ $$


For $a=0,$ $I=0$

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    $\begingroup$ In your first display equation: Outer "$a$" should be $|a|$ because $\sqrt{a^2} = |a|$ ($\sqrt{\phantom{a}}$ only produces nonnegative output). Inner "$a$" should be $a^2$. $\endgroup$ Commented Feb 8, 2016 at 13:44
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    $\begingroup$ @EricTowers, in first display it is fine, as $a>0$, thus $|a|=a$. $\endgroup$
    – Galc127
    Commented Feb 8, 2016 at 17:03
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    $\begingroup$ @Galc127 : Don't know how I missed that $a$ was given positive in the OP. I wonder if it was edited in later... $\endgroup$ Commented Feb 8, 2016 at 23:52
  • $\begingroup$ You should change the limits of integration when the variable is $t$ as when $x=a$, $t=1$. $\endgroup$
    – N74
    Commented Feb 9, 2016 at 8:19
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Notice, $$\int_0^a\frac{1}{\sqrt{a^2-x^2}}\:\mathrm dx$$ Let $x=a\sin\theta\implies \mathrm dx=a\cos\theta\:\mathrm d\theta$, \begin{align}&=\int_{0}^{\pi/2}\frac{1}{a\cos \theta}(a\cos\theta\ \, \mathrm d\theta)\\ &=\int_{0}^{\pi/2}\,\mathrm d\theta \\& =\left(\theta\right)_{0}^{\pi/2}\\&=\frac \pi 2\end{align}

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    $\begingroup$ wow nice. Good method as well $\endgroup$
    – RStyle
    Commented Feb 8, 2016 at 12:40
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Different method here.

Consider $$\dfrac{d}{da} \int_0^a \sqrt{a^2-x^2} dx = a \int_0^a \frac{1}{\sqrt{a^2-x^2}} dx + \underbrace{\sqrt{a^2-a^2}}_0$$ by the Leibnitz rule of differentating under the integral.

Now, the integral on the left-hand side is just a quarter of the area of the circle of radius $a$ (because the integrand is the $y$-coordinate of the upper-right quadrant of the circle at $x$-coordinate $x$), so the LHS is $$\dfrac{d}{da} \frac{\pi a^2}{4} = \frac{\pi a}{2}$$

Therefore the required integral is $$\frac{\pi}{2}$$

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