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I was thinking about the following integral if I could solve it without using trigonometric formulas. If there is no other way to solve it, could you please explain me why do we replace $x$ with $2\sqrt 2 \sin(t)$? I'm really confused about these types of integrals.

$$\int \sqrt{8 - x^2} dx$$

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  • $\begingroup$ I'm actually confused how is that the integral is related to trigonometric formulas... $\endgroup$ – Pichi Wuana Feb 8 '16 at 19:41
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    $\begingroup$ @PichiWuana, what shape is the curve $\sqrt{8-x^2}$? $\endgroup$ – jameselmore Feb 8 '16 at 20:35
  • $\begingroup$ @jameselmore I saw a graph in WolframAlpha. It could be said that it has somehow a shape of $\cos x$...right? $\endgroup$ – Pichi Wuana Feb 10 '16 at 15:22
  • $\begingroup$ @PichiWuana, $y^2 + x^2 = \sqrt 8^2 \implies y = \pm \sqrt{8 - x^2}$. It's the top half of a circle $\endgroup$ – jameselmore Feb 10 '16 at 15:31
  • $\begingroup$ @jameselmore Oh... okay. Thanks! $\endgroup$ – Pichi Wuana Feb 10 '16 at 18:21
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By rescaling the variable, let us replace the constant $8$ by $1$, for convenience.

The equation $y=\sqrt{1-x^2}$ represents the upper-half of the unit circle, and the integral

$$\int_{t=0}^x\sqrt{1-t^2}dt$$ is the area of a vertical "slice" between the abscissas $0$ and $x$. You can compute it as the area of a sector of aperture $\theta$ such that $\sin(\theta)=x$, plus a triangle of base $x$ and height $\sqrt{1-x^2}$.

enter image description here

Hence,

$$A=\frac12\theta+\frac12x\sqrt{1-x^2}=\frac12\arcsin(x)+\frac12x\sqrt{1-x^2}.$$

This is how a trigonometric function appears, and you can't avoid it because it belongs to the final solution.


You also see the connection by taking the derivative

$$(\arcsin(x))'=\frac1{\sqrt{1-x^2}}.$$ The trigonometric function disappears and is replaced by a rational expression.

A similar phenomenon occurs with the logarithm,

$$(\ln(x))'=\frac1x,$$

and this is why you will see logarithms appear now and then in antiderivatives.

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The idea is that we want to get rid of the square root. So we use the fact that $1 - \sin^2 = \cos^2$, so that $$\sqrt{1 - \sin ^2 x } = \sqrt{\cos^2 x} = |\cos x|$$

which we then can integrate (of course there will be another factor coming from $dx$ but that works out)

Same idea when you're faced with $\sqrt{1 + x^2}$; this time we use the fact that $1 + \sinh^2 = \cosh^2$ and we get rid of the square root in the same way.

Of course if you have a number $a$ instead of $1$, you need to be able to factor that; so you want to transform $\sqrt{a - x^2}$ in $\sqrt{a - a\sin^2 x} = \sqrt a \sqrt{1 - \sin^2 x} =\sqrt a |\cos x|$

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    $\begingroup$ The OP asks if it can be done without using a trig substitution. $\endgroup$ – zz20s Feb 8 '16 at 19:37
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    $\begingroup$ A little correction :$\sqrt{\cos^2 x} = |\cos x|.$ $\endgroup$ – C. Dubussy Feb 8 '16 at 19:38
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    $\begingroup$ @zz20s He also asked the idea behind the use of trigonometric formulas $\endgroup$ – Ant Feb 8 '16 at 19:47
  • $\begingroup$ @C.Dubussy Thanks! Corrected :) $\endgroup$ – Ant Feb 8 '16 at 19:48
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Let
$$I=\int\sqrt{8-x^2}\ dx\tag 1$$ using integration by parts,
$$I=\sqrt{8-x^2}\int 1\ dx-\int \left(\frac{-2x}{2\sqrt{8-x^2}}\right)\cdot x\ dx$$ $$I=\sqrt{8-x^2}(x)-\int \frac{(8-x^2)-8}{\sqrt{8-x^2}} \ dx$$ $$I=x\sqrt{8-x^2}-\int \left(\sqrt{8-x^2}-\frac{8}{\sqrt{8-x^2}} \right)\ dx$$ $$I=x\sqrt{8-x^2}-\int\sqrt{8-x^2}\ dx+8\int \frac{1}{\sqrt{8-x^2}}\ dx$$ setting the value from (1), $$I=x\sqrt{8-x^2}-I+8\int \frac{1}{\sqrt{(2\sqrt 2)^2-x^2}}\ dx$$ $$2I=x\sqrt{8-x^2}+8\sin^{-1}\left(\frac{x}{2\sqrt 2}\right)+c$$ $$I=\color{red}{\frac{1}{2}\left(x\sqrt{8-x^2}+8\sin^{-1}\left(\frac{x}{2\sqrt 2}\right)\right)+C}$$

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    $\begingroup$ The solution used trig substitution in the last integral anyway. It has to use trig substituiton to integrate this type of integrand. $\endgroup$ – runaround Feb 8 '16 at 19:54

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