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.

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

  • $\begingroup$ 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

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



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,


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


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

$$I=\int\sqrt{8-x^2}\ dx\tag 1$$ 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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