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What does $ d\tan(x) = \sec^2(x)\,dx$ mean?

I've seen it used in integration problems to make it more simpler. However, I'm not really sure what it means. Can someone explain this to me?

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  • $\begingroup$ Could you clarify which part you're having trouble with? $\endgroup$ – Mike May 7 '13 at 18:13
  • $\begingroup$ When you substitute in dtan(x) into a problem, it gets rid of the dx which indicates which variable to integrate with respect to. $\endgroup$ – Bob Shannon May 7 '13 at 18:14
  • $\begingroup$ How would this identity be used to integrate $ sec^2(x) dx $? $\endgroup$ – Bob Shannon May 7 '13 at 18:14
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It's really just shorthand for writing out the full variable replacement. Instead of writing: $$\int \sec^2(x)dx= \int \left(\frac{d}{dx} \tan(x)\right)dx= \tan(x)+C$$ You can just write: $$\int \sec^2(x)dx= \int d\tan(x)$$ Which is just like writing:

$$\int dy = y+C \ \to \ \int d\tan(x) = \tan(x)+C$$

As your example in the comments suggests this works for more complex integrals as well:

$$\int \sec^2(x)\tan(x)dx = \int \tan(x)\ d\tan(x) = \frac{\tan^2(x)}{2}+C$$ Much like: $$\int y dy = \frac{y^2}{2}+C $$

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  • $\begingroup$ It's saying that the integral of sec^2(x) = tan(x)? $\endgroup$ – Bob Shannon May 7 '13 at 18:19
  • $\begingroup$ yep (up to a constant of course) $\endgroup$ – nbubis May 7 '13 at 18:21
  • $\begingroup$ Would this substitution be useful for integrating something like $ sec^2(x)tan(x) dx $? Or would we still need to use integration by parts? $\endgroup$ – Bob Shannon May 7 '13 at 18:22
  • $\begingroup$ @Bob - see my updated answer. $\endgroup$ – nbubis May 7 '13 at 18:30
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Differentials measure the tendency of a function to change given an infinitesimally small change of the dependent variable. The expression $d\tan x=\sec^2 xdx$ means that the difference $\tan x_1-\tan x_0$ is approximately $\sec^2x_0\cdot(x_1-x_0)$ when $x_1$ is near $x_0$.

In other words, if we reduce to infinitesimal changes of $x$, we are saying that the slope of the function $\tan x$ at a point $x_0$ is $\sec^2x_0$.

As others have already noted, the value of expressions like this in computing integrals is that it is often easier to see an antiderivative having changed variables. In some sense, we are allowing $\tan x$ to become our new independent variable in the integration.

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I don't know if it's common practice, but when I was taught calculus that notation for differentiation as sort of an implicit method of using the chain rule. For example

$$d(\sin(xe^x))=\cos(xe^x)d(xe^x)=\cos(xe^x)(e^xdx+xd(e^x))=\cos(xe^x)(xe^x+e^x)dx$$

If it makes you more comfortable, replace $d()$ with $\frac {d()}{dx}$.

$$\frac{d(\sin(xe^x))}{dx}=\cos(xe^x)\frac{d(xe^x)}{dx}=\cos(xe^x)(e^x+x\frac{de^x}{dx})=\cos(xe^x)(xe^x+x)$$

or in your case

$$\frac{d(\tan x)}{dx}=\sec^2x$$

Or if you wish to embrace the notation

$$u=\tan x,du=d(\tan x)=\sec^2xdx$$

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