I have learned about the method of integrating functions involving another function and its derivative by using integration by substitution. For example $$\int 5x^4 e^{x^5}~dx$$ In the case, let $$u(x) = x^5$$ $$u'(x)= 5x^4$$ $$v(x) = e^x$$ Therefore, the integral can be expressed as $$\int u'(x)v(u(x))~dx = V(u(x))$$ where $V$ is an antiderivative of $v$. This is how I interpreted integration by substitution when I first saw it. However, most of the resources found the derivative of $u$ and substituted the value of $dx$ that they got in terms of $du$ by "pretending" that $\dfrac{du}{dx}$ is a fraction. I didn't like how they did that so I continued with my interpretation of the method. But when I read about trigonometric substitution, I again saw $\dfrac{d \theta}{dx}$ being treated as a fraction and using it to find an expression for $dx$ in terms of $du$. Since u- substitution is the counterpart of chain rule in differentiation, I was able to understand how treating the derivative as a fraction "worked" but I can't seem to figure out how it works for trigonometric substitution.
Why can we treat $\dfrac{d \theta}{dx}$ as a fraction while using trigonometric substitution?
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0When any book/ professor/ that-one-crazy-hobo-who-lives-next-to-Racetrac says something like, "because $\frac{dy}{dx} = \text{blah}$, we can say that $dy = \text{blah}\, dx$", they're just being informal because it works (most of the time) and many students (like engineers) don't really need the rigorous explanation for *why* integration by substitution works. But the actual theorem goes the way you've already expressed. So just keep interpretting it the correct way and you should be fine. – 2017-02-21
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0... Unless you're saying that you tried reinterpretting some $d\theta = \text{blah}\,dx$ problem as an integral involving $\frac{d\theta}{dx}$ and it didn't work. In which case I'll need to see an example of it not working to say what's going on. – 2017-02-21
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0@Bye_World Typical introductory questions to trigonometric substitution like finding the integral $\int \dfrac{1}{\sqrt{4 - x^2}} ~dx$. We assume $$x = 2 \sin \theta$$ $$dx = 2 \cos \theta d \theta$$ – 2017-02-21
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0Ah. So you're not understanding the difference between a pullback and a pushforward. Hold up, I've answered a similar question before. Let me find it. – 2017-02-21
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0What is one example where we cannot treat $dy/dx$ as a fraction? If $y = \sin(x)$, then $dy = \cos(x) dx$. It is working machinery, and the $dx$ is already present, so why not? – 2017-02-21
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0@Kaynex I agree that it **can** be treated as a fraction. But again, it is not really a fraction but a limit. – 2017-02-21
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2@Parth Try reading through my answers [here](http://math.stackexchange.com/questions/1575554/understanding-integration-by-change-of-variables/1575569) and [here](http://math.stackexchange.com/questions/1506621/the-inegral-int-12-2x-sqrtx2-1-dx-using-differential-forms/1508443) and let me know if that helps. – 2017-02-21
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0@Parth It's really both a limit and a fraction. $dy/dx$ represents the best fitting tangent line, but that is also a fraction between infinitesimal elements. The chain rule agrees that these elements are multiplicative. – 2017-02-21
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0@Bye_World Thanks for the links. I was able to understand why we use trigonometric functions to simplify integrals by the help of the first link. It was very helpful. – 2017-02-21
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0No problem. :-) – 2017-02-21
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0@Bye_World Your answers made it clear how to find definite integral of such functions but what about the indefinite integral? – 2017-02-23
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0@Parth Using the fundamental theorem of calculus, you can show that substitutions work almost exactly the same way for indefinite integrals. The only difference is that you can't change your limits of integration so you have to change your variables back to the original ones at the end. – 2017-02-26