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I found the following integral in a calculus textbook.

$$\int_{-1}^1\frac{\mathrm d x}{\sqrt{4-x^2}}$$

What is $\mathrm d x$ doing up there and how do I fix this??? I didn't even know it was possible to write an integral like this. Much help needed.

Rodrigo de Azevedo
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3 Answers3

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The $dx$ is a symbol, that tells you by which variable to integrate. But it is often treated like a factor. Different ways to write this integral would be $$ \int_{-1}^1 \frac{dx}{\sqrt{4-x^2}} = \int_{-1}^1 \frac{1}{\sqrt{4-x^2}}dx = \int_{-1}^1 dx \frac{1}{\sqrt{4-x^2}}. $$ The first variant is often used to save space, the last variant is often used in a physics context.

In order to solve this, we can use a substitution. $\sqrt{4-x^2}$ has actually something to do with a circle, so we choose a substitution that involves trig functions. In this case $x=2\sin t$ would be a good choice. We see that for $t=-\frac\pi6$ it is $-1=2\sin(-\frac\pi6)$ and for $t=\frac\pi6$ it is $1=2\sin(\frac\pi6)$. Also $2\sin t$ is monotonous on the interval $(-\frac\pi6,\frac\pi6)$. We have done this to find out the limits of our substituted integral.

Now we need the derivative of $x$ with respect to $t$: $$ \frac{dx}{dt} = 2\cos t.$$ Since we want to substitute $t$ for $x$, we treat $dx$ and $dt$ symbolically as numbers in order to solve for $dx = 2\cos t\;dt$.

Now we plug everything together: \begin{align*} \int_{-1}^1\frac{dx}{\sqrt{4-x^2}} &= \int_{-\frac\pi6}^{\frac\pi6} \frac{2\cos t\;dt}{\sqrt{4 - (2\sin t)^2}} \\ &= \int_{-\frac\pi6}^{\frac\pi6} \frac{2\cos t\;dt}{\sqrt{4(1 - \sin^2 t)}} \\ &= \int_{-\frac\pi6}^{\frac\pi6} \frac{2\cos t\;dt}{2\sqrt{\cos^2 t}}, \end{align*} where we have used the trigonometric pythogoras $1=\sin^2+\cos^2$. Now we see that for $t\in(-\frac\pi6,\frac\pi6)$ we have $\cos t \geq 0$ and thus $\sqrt{\cos^2t} = \cos t$. Going on: \begin{align*} \int_{-1}^1\frac{dx}{\sqrt{4-x^2}} &= \int_{-\frac\pi6}^{\frac\pi6} \frac{2\cos t\;dt}{2\sqrt{\cos^2 t}} \\ &= \int_{-\frac\pi6}^{\frac\pi6} \frac{\cos t\;dt}{\cos t} \\ &= \int_{-\frac\pi6}^{\frac\pi6} dt = \int_{-\frac\pi6}^{\frac\pi6} 1 dt \\ &= [t]_{-\frac\pi6}^{\frac\pi6} = \frac\pi6 - (-\frac\pi6) = 2\frac\pi6 = \frac\pi3. \end{align*}

Stefan Hante
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  • Okay thank you but I still need help to solve the rest –  Oct 27 '17 at 08:20
  • You should show your work so far then. What did you try? Where are you stuck? – Stefan Hante Oct 27 '17 at 08:22
  • I assume that $(4-x^2)^{-.5}$ is some kind of circle and then I can use the area of a circle to calculate the rest but the best I got was $2^2=y^{-2}+x^2$ and then I got stuck –  Oct 27 '17 at 08:24
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    There is good reason for $dx$ to be treated like a factor. It is not merely a symbol. See this: https://math.stackexchange.com/questions/200393/what-is-dx-in-integration/200403#200403 – Michael Hardy Oct 30 '17 at 23:15
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$$\int_{-1}^{1}\dfrac{dx}{\sqrt{4-x^2}}$$ is the same as $$\int_{-1}^{1}\dfrac{1}{\sqrt{4-x^2}}dx$$From there, you solve it as you normally would.

user
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The presence of $dx$ above the fraction line is not some sort of new notation. Now, for solving this. Let $$x=2\sin u\Rightarrow dx=2\cos u\\ -1=2\sin u_1, 1=2\sin u_2\\u_1=\arcsin(-1/2), u_2=\arcsin(1/2) \Rightarrow\\\int \frac{1}{\sqrt{4-x^2}}dx=\int \frac{1}{\sqrt{4-4\sin^2 u}}2\cos u du=\\\int \frac{1}{2\sqrt{1-\sin^2u}}2 \cos u du=\int\frac{1}{2{\cos u}}2\cos udu=\int du$$

So $$I=\int_{u_1=\arcsin(-1/2)}^{u_1=\arcsin(1/2)}du=\arcsin(1/2)-\arcsin(-1/2)=2\arcsin(1/2)=1.047$$

$($we made use of the fact that $\arcsin(-x)=-\arcsin(x)$ on the last step$)$

MathematicianByMistake
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