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I encountered the below formula in my text.
I know the author is using integration by substitution and double angle formula
but for some reason, every working on paper that i did is different from the answer below:

$\int_{-a}^{a}2(\sqrt{a^2-x^2})dx $

Can anyone help me to expand the above equation.
I can't wrap my head on how above answer is reproduced.

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The answer given above seems to be wrong, and even evaluating, it is off by a factor of 2. – robjohn Oct 1 '12 at 23:06
there's a bracket after $\frac{1}{2}$ that cover both terms of the equation.In fact it matches with the workout (ie: (4)) you gave below. – kypronite Oct 2 '12 at 2:36
The bracket appears to be before the $\frac12$, so the answer doesn't appear to match. There is also an $x$ missing that should be multiplying $\sqrt{a^2-x^2}$ – robjohn Oct 2 '12 at 2:38
I have a look again at your solution below.You are right, the numbers in my original post are out of place. – kypronite Oct 2 '12 at 3:11
up vote 5 down vote accepted

This can be done with a standard $\sin$ substitution: $(1)\ x=a\sin(\theta)$ followed by a change of variables: $(2)\ \phi=2\theta$. $$ \begin{align} \int_{-a}^a2\sqrt{a^2-x^2}\,\mathrm{d}x &=\int_{-\pi/2}^{\pi/2}2a\cos(\theta)\,\mathrm{d}a\sin(\theta)\tag{1}\\ &=2a^2\int_{-\pi/2}^{\pi/2}\cos^2(\theta)\,\mathrm{d}\theta\\ &=2a^2\int_{-\pi/2}^{\pi/2}\frac{1+\cos(2\theta)}{2}\,\mathrm{d}\theta\\ &=2a^2\int_{-\pi}^\pi\frac{1+\cos(\phi)}{4}\,\mathrm{d}\phi\tag{2}\\ &=2a^2\left[\frac{\phi+\sin(\phi)}{4}\right]_{\phi=-\pi}^{\phi=\pi}\\ &=\pi a^2 \end{align} $$ However, the answer you show in the question looks like the answer that comes from an integration by parts: $u=\sqrt{a^2-x^2}$ and $\mathrm{d}v=\mathrm{d}x$ so that $v=x$ and $\mathrm{d}u=-\frac{x\,\mathrm{d}x}{\sqrt{a^2-x^2}}$ $$ \begin{align} \int2\sqrt{a^2-x^2}\,\mathrm{d}x &=2x\sqrt{a^2-x^2}+\int\frac{2x^2}{\sqrt{a^2-x^2}}\,\mathrm{d}x\\ &=2x\sqrt{a^2-x^2}-\color{#C00000}{\int\frac{2(a^2-x^2)}{\sqrt{a^2-x^2}}\,\mathrm{d}x}+\int\frac{2a^2}{\sqrt{a^2-x^2}}\,\mathrm{d}x\tag{3} \end{align} $$ Adding the integral in red to both sides of $(3)$ and dividing by $2$ yields $$ \begin{align} \int2\sqrt{a^2-x^2}\,\mathrm{d}x &=\frac12\left[2x\sqrt{a^2-x^2}+\int\frac{2a^2}{\sqrt{a^2-x^2}}\,\mathrm{d}x\right]\\ &=\frac12\left[2x\sqrt{a^2-x^2}+2a^2\,\sin^{-1}(x/a)\right]+C\tag{4} \end{align} $$ Evaluating $(4)$ at the limits of integration yields $$ \begin{align} \int_{-a}^a2\sqrt{a^2-x^2}\,\mathrm{d}x &=\left[x\sqrt{a^2-x^2}+a^2\,\sin^{-1}(x/a)\right]_{-a}^a\\ &=\pi a^2 \end{align} $$ Not quite what you got, but of the same form.

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Oh, well, didn't see you posted an answer too. – Pedro Tamaroff Oct 2 '12 at 0:59
actually the equation I gave in my original post is incomplete for clarity reason. – kypronite Oct 2 '12 at 2:32

Step-by-step solution from WolframAlpha

$$2\int_{ - a}^a {\sqrt {{a^2} - {x^2}} dx} = 4\int_0^a {\sqrt {{a^2} - {x^2}} dx} $$

$$\eqalign{ & x = a\sin t \cr & dx = a\cos t \cr} $$

$$x\to a\Rightarrow t\to\pi /2\\x\to 0\Rightarrow t\to 0$$

$$\eqalign{ & = 4a\int_0^{\pi /2} {a\sqrt {1 - {{\sin }^2}t} \cos tdt} \cr & = 4{a^2}\int_0^{\pi /2} {{{\cos }^2}tdt} \cr & = 4{a^2}\int_0^{\pi /2} {\frac{{1 + \cos 2t}}{2}dt} \cr & = 4{a^2}\left. {\frac{{2t + \sin 2t}}{4}} \right|_0^{\pi /2} = 4{a^2}\frac{\pi }{4} = \pi {a^2} \cr} $$

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thanks, i know about step by step solution from wolfram alpha but didn't realize how useful is it until now...thanks – kypronite Oct 1 '12 at 18:30
It would be nice to bring the answer here. The link provided no longer seems to work. – robjohn Oct 1 '12 at 21:53
@robjohn Thy will be done. – Pedro Tamaroff Oct 2 '12 at 0:58

let $x = a \sin(u)$ then use $\sin^2(u) + \cos^2(u) = 1$

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