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Calculate $$\int \dfrac{1}{\sqrt{4-x^2}}dx$$

Suppose that I only know regular substitution, not trig.

I tried to get help from an integral calculator, and what they did was:

$$\text{Let u = $\frac{x}{2}$} \to\dfrac{\mathrm{d}u}{\mathrm{d}x}=\dfrac{1}{2}$$

Then the integral became:

$$={\displaystyle\int}\dfrac{1}{\sqrt{1-u^2}}\,\mathrm{d}u = \arcsin(u) = \arcsin(\frac{x}{2})$$ And I'm not sure how they accomplished this, where did the 4 go? I understand the arcsin part but not sure how they got rid of the 4? Also how did they know to substitute $\frac{x}{2}$? It doesn't seem very obvious to me.

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    Hint : $\sqrt{4-x^2}=2\sqrt{1-\frac{x^2}{4}}$ – 2017-02-02

4 Answers 4

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Recall that $\frac{d}{dx}(arcsin x) =\frac{1}{\sqrt{1-x^2}}$ and hence posing $u=\frac{x}{2}$ and hence $dx=2du$ substituting you have $ \int \frac{du}{\sqrt{1-u^2}} = \arcsin u = \arcsin \frac{x}{2} $

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Since $\text{d}x=2\text{d}u$, we have $2$ both in the numerator and in the denominator and it reduces.

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$$\int \frac{\text{d}x}{\sqrt{4-x^2}}=\int \frac{2 \ \text{d}u}{\sqrt{4-(2u)^2}}=\int \frac{2 \ \text{d}u}{\sqrt{4(1-u^2)}}=\int \frac{2 \ \text{d}u}{2\sqrt{1-u^2}}=\int \frac{ \text{d}u}{\sqrt{1-u^2}} $$

Why especially this substitution: Notice that

$$\int \frac{\text{d}x}{\sqrt{4-x^2}}=\int \frac{\text{d}x}{\sqrt{4\left(1-\frac14x^2\right)}}=\int \frac{\text{d}x}{2\sqrt{1-\left(\frac{x}{2} \right)^2}}$$

so you can see that it is quite nice to substitute $u=\frac{x}{2}$; we get a function $\frac{1}{\sqrt{1-u^2}}$ and we already know the integral to this one.

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    How did they know to substitute x/2 though? – 2017-02-02
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    @KSplitX I edited my post so you can see why we set $u$ to $x/2$. I hope it helps. – 2017-02-02
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$\int \dfrac{1}{\sqrt{4-x^2}}dx=\int \dfrac{\frac{1}{2}}{\frac{\sqrt{4-x^2}}{2}}dx=\int \dfrac{\frac{1}{2}}{\sqrt{{\frac{4}{4}-(\frac{x}{2})^2}}}dx=\int \dfrac{\frac{1}{2}}{\sqrt{{1-(\frac{x}{2})^2}}}dx=arcsin (\frac{x}{2}) + C$.