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How in indefinite integral we can do the substitution without bothering about the variable or function we are substituting the variable with? For example, it is very common to make trigonometric substitutions for evaluating indefinite integrals like we put x = sint or cosect or cost etc. But how can we do that when we know that the original variable x can assume any real values but the function sint, cost by definition can only asse values in [-1,1] . Now It is okay to make the substitutions x in terms of tan or cot as they can assume all real values. But how is it justified for other trig functions? Also why in indefinite integrals we generally ignore the modulus or domain etc. As in Paul's online notes http://tutorial.math.lamar.edu/Classes/CalcII/TrigSubstitutions.aspx While solving and only take care in definite integral?

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    I don't exactly know about others, but when I do trig subs I also solve for $\theta$ so when I get to my final answer I can plug the original variable back in, eliminating any issue of switching limits, etc. This is how we can do trig subs with indefinite integrals. – 2017-02-28
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    I know how to do it, i am asking how is it correct to assume a given arbitrary variable equal to another trig function whose range is not same as that of original variable ? – 2017-02-28
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    We only use $\sin$ or $\cos$ substitutions when the range of the part we substituted is [-1,1]. Most of the times we will not claim the range is between $-1$ and $1$, and that might be the reason why we feel the variable can be any real number. – 2017-02-28
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    @Matt, Hey, I have all these doubts now too. For a moment I wondered whether I had asked this question! I want to know whether you have had these doubts cleared since you asked this and if yes, can you help me too -- by explaining it or providing helpful links or references, etc.? – 2018-06-13
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    @MrReality I think the provided answer pretty much clears my basic doubt. Basically, any good mathematical text will always ensure that whenever substitution is made by sin, cos, sec or cosec then always a constant will be multiplied by it which depends on the expression to which trig quantity is substituted in the indefinite integral. For example, though the range of sinx is [-1,1] but that of a.sinx is [-a,a]. – 2018-06-13
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    @MrReality Hence whenever we want to make a substitution where required range is all real we use tan or cot but when we want any other range where there are some values of x where the integrand is not defined we use other trig ratios. – 2018-06-13
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    @MrReality https://en.wikipedia.org/wiki/Trigonometric_substitution – 2018-06-13
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    @MrReality If you have any particular doubt or example you are stuck on. DO ask. Happy to help :) – 2018-06-13
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    @MrReality Note one important thing in wiki article. It initially states to use asinx as substituition in the example only when the expression is in square root whereas when in third example in expression containing no square root it explicitly stated to solve this integral by using partial fractions rather than trig substitution because that substitution cannot cover all values of x and hence is invalid. It requires a square root to be present. – 2018-06-13
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    @Matt, thanks for replying to this : ). I have had these doubts for a year in which I studied integral calculus but didn't ask anybody to clarify them, as I thought they were too basic/stupid since my Math teacher hadn't clarified them in class to us. I did guess that that may be why me multiply a constant to sin and cos functions, but certain solutions confused me regardless. I didn't keep track of those questions, but recently I am re-learning bits of Math I ignored in the last couple years, so if I take up integration and reach those questions again ... [continued] – 2018-06-14
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    [continued]... and still have the confusion, I'll ask you, if that's okay. What would you prefer: me tagging you in a question I post on this site, or in this question's comments, or using some other way? – 2018-06-14
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    @Matt, I couldn't find the rule about choosing a value in the principal value branch of the corresponding trig function to substitute an angle in trig substitution in a definite integral in the link: https://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Principal_values you gave. If it's not in it, is there another link which does mention it? – 2018-06-14
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    @MrReality Sure, you can ask me. Just tag me in the comments. imo no doubt is stupid. As fir the link I am not sure what you’re asking but look at the table in the link it states the ranges of angles in principal value branch for all six trig ratios. Could you be more clear as to what you’re looking for! – 2018-06-16

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Let's consider the first use of trigonometric substitution on the page you linked: $$ \int \frac{\sqrt{25x^2 - 4}}{x} \,dx. $$

Observe that the integrand is not real for all values of $x.$ For example, if you put $x=\frac15,$ the integrand becomes $$ \frac{\sqrt{25\left(\frac15\right)^2 - 4}}{\left(\frac15\right)} = 5 \sqrt{-3}. $$

In fact, the integrand is defined only when $25x^2 - 4 \geq 0,$ that is, for $x \in \left(-\infty,-\frac25\right] \cup \left[\frac25,\infty\right).$ Notice that the recommended substitution in this integral is $$ x = \frac25 \sec\theta, $$ which happens to be able to take on any value in $\left(-\infty,-\frac25\right] \cup \left[\frac25,\infty\right).$

Later on the same page, there is an integral in which the expression $\sqrt{9 - x^2}$ occurs. As explained in that example, the $\sec\theta$ substitution does not work in that case, because it results in taking a square root of a negative value (except at some isolated points). Instead, we substitute $$ x = 3 \sin\theta, $$ which very happily can produce all values of $x$ in the interval $[-3,3],$ which happens to be all the values of $x$ for which $9 - x^2 \geq 0.$

In short, the premise that $x$ can take on every real value in every indefinite integral is not true. There can be large gaps in the set of values of $x$ for which the integrand is defined over real numbers. In some cases $x$ can only take values in a finite interval. The typical trigonometric substitutions that we make for $x$ do not always support every real number as a value of $x,$ but they are able to produce all values of $x$ for which the integral is defined.

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    I share the OP's doubt. I realise this is slightly unrelated to the points in your answer but still can you explain why in the solution of the second problem on the link OP gave we take $\theta = 0$ & $\pi/3 $, instead of $0$ & $5\pi/3$ *or* $2\pi$ & $5\pi/3$? Is there some rule regarding this that says we have to consider the values belonging to a particular domain for such integration problems? – 2018-06-13
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    @MrReality Yes. There is for this we take particular values according to the range of corresponding inverse trigonometric function. For example, we have sin(x) = a; then x = sin inverse(a) = a value in [-pi,pi] whereas for cos we have [0,pi]. This is called Principal value. This is just a convention and requires that each such interval must cover entire range of trig function ie [-1,1] here. See this : – 2018-06-13
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    @MrReality https://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Principal_values – 2018-06-13
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    @MrReality The integral after substitution still must integrate the same “thing”, although it’s allowed to use Riemann sums (or whatever sums your theory of integration is based on) that slice up that “thing” differently than the original integral. If we let $\theta$ go from $0$ to $5\pi/3,$ then $x=\frac25\sec\theta$ would make $x$ go through all the numbers from $\frac25$ upward, then go through all the numbers less than $-\frac25$ twice (first going up, then down), and finally come back down to $\frac45$. This is not a simple integral. Why would we want to mess with that? – 2018-06-13
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    @DavidK, thanks for answering my doubt. So, 0 & 5π/3 would get us an incorrect answer, right? But, why didn't we choose 2π & 5π/3 then? Is there actually a rule that when doing trig substitutions in definite integrals, the angles that are substituted should lie within a particular branch, (such as the principal value branch of the corresponding inverse trig function)? Or is it just a matter of choice and/or convenience? – 2018-06-14
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    Actually you _could_ choose $2\pi$ and $5\pi/3.$ You would have to be careful about positive/negative signs (that integral would go from the larger number to the smaller), but if you can do every step correctly I believe it works out. Even from $0$ to $5\pi/3,$ although it is an improper integral, I think the Cauchy principal value comes out to the correct answer. But the point of doing a substitution is usually to make the integral easier to solve, so we tend to use smaller, more familiar angles in trig substitution and avoid improper integrals. – 2018-06-15
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    @DavidK, thanks. Sorry to bother again but --just to confirm -- there is, in fact, **no rule** which dictates using only the angles falling in the principal value branch of the particular trigonometric function -- tan, cosec, anything? It is just a matter of convenience? I was so confused because I had only seen the principal values used in *all* the integration problems.. – 2018-06-15
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    @MrReality No rule I know of, just convention and convenience. – 2018-06-15
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    @DavidK, thank you so much! – 2018-06-16