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I am having a little problem with my maths homework. The problem is as follows:

\begin{equation} \int^a_0{\cfrac{dx}{x \ + \ \sqrt{a^2 \ - \ x^2}}} \end{equation}

I tried to do the following but got stuck halfway:

Let $\ \ x \ = asin\theta, \ hence, \ dx = acos\theta \ d\theta $

$ \int^a_0{\cfrac{dx}{x \ + \ \sqrt{a^2 \ - \ x^2}}} $

$ = \int^\frac{\pi}{2}_0{\cfrac{acos\theta}{asin\theta \ + \ \sqrt{a^2 \ - \ a^2sin^2\theta}}}\ d\theta $

$ = a \cdot \int^\frac{\pi}{2}_0{\cfrac{(cos\theta \ + \ sin\theta) \ + \ (cos\theta \ - \ sin\theta) - \ cos\theta}{asin\theta \ + \ \sqrt{a^2cos^2\theta}}}\ d\theta $

$ = a \cdot \int^\frac{\pi}{2}_0{\cfrac{(cos\theta \ + \ sin\theta) \ + \ (cos\theta \ - \ sin\theta) - \ cos\theta}{asin\theta \ + \ acos\theta }}\ d\theta $

$ = \int^\frac{\pi}{2}_0{\cfrac{(cos\theta \ + \ sin\theta) \ + \ (cos\theta \ - \ sin\theta) - \ cos\theta}{sin\theta \ + \ cos\theta }}\ d\theta \\ \\ $

$ = \int^\frac{\pi}{2}_0{\left(1 \ + \ \cfrac{(cos\theta \ - \ sin\theta)}{sin\theta \ + \ cos\theta } - \cfrac{cos\theta}{sin\theta \ + \ cos\theta}\right)}\ d\theta $

Could someone please advise me how to solve this problem?

Donald
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3 Answers3

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$$I=\int^a_0{\cfrac{dx}{x \ + \ \sqrt{a^2 \ - \ x^2}}}$$

Let $x^2+y^2=a^2\implies dx=-\frac{y}{\sqrt{a^2-y^2}}dy$

Thus:

$$I=\int^a_0{\cfrac{dy}{y \ + \ \sqrt{a^2 \ - \ y^2}}}\frac{y}{\sqrt{a^2-y^2}}$$

Adding these together,

$$2I=\int^a_0{\cfrac{dx}{x \ + \ \sqrt{a^2 \ - \ x^2}}}\left(1+\frac{x}{\sqrt{a^2-x^2}}\right)=\int_0^a\frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\left(\frac{x}{a}\right)\vert_0^a=\frac{\pi}{2}$$

Thus, $I=\frac{\pi}{4}$.

πr8
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  • May I know how is it possible to substitute y = x into your last equation? $ y^2 = a^2 - x^2 $ right? I am referring to this part: $ \left(1+\frac{x}{\sqrt{a^2-x^2}}\right) $ – Donald Sep 15 '16 at 10:26
  • It's because $y$ is just a dummy variable - $\int_0^a f(x),dx = \int_0^a f(y),dy = \int_0^a f(t),dt = \cdots$ etc. So you just put everything in terms of $x$ and add them together. – πr8 Sep 15 '16 at 10:29
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Assuming $a>0$ and applying the substitutions $x=az$, $z=\sin\theta$: $$ I(a)=\int_{0}^{a}\frac{dx}{x+\sqrt{a^2-x^2}}=\int_{0}^{1}\frac{dz}{z+\sqrt{1-z^2}}=\int_{0}^{\pi/2}\frac{\cos\theta}{\sin\theta+\cos\theta}\,d\theta $$ but due to the substitution $\theta=\frac{\pi}{2}-\varphi$ we also have $I(a)=\int_{0}^{\pi/2}\frac{\sin\theta}{\sin\theta+\cos\theta}\,d\theta$, hence: $$ 2\cdot I(a) = \int_{0}^{\pi/2}\frac{\sin\theta+\cos\theta}{\sin\theta+\cos\theta}\,d\theta = \frac{\pi}{2}$$ and $\boxed{I(a)=\color{red}{\large\frac{\pi}{4}}}$ holds by symmetry.

Jack D'Aurizio
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Try Euler substitutions. The second one might do the job.

9cvele3
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