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So I was solving this question: $$\int{1+\tan(x)\tan(x+ \alpha)} \ dx$$ In the process of solving, I had to evaluate the integration of $tan(x)$ Which I did this way:
$\int{\tan(x)} \ dx=\int{\frac{\sin(x)}{\cos(x)}} \ dx=$$ -\int{\frac{1}{\cos(x)}d(\cos(x))}=-\ln|\cos(x)|+C$

I did it this way
$\int {uv} \ dx = u \int{v} \ dx - \int({u'} \int{v} dx) dx $ but something went wrong: $$Let \int{\tan(x)} \ dx= \int {\frac{\sin(x)}{\cos(x)}} \ dx = I(x)$$ $$\implies I(x)=\int{\sin(x) \sec(x)}dx$$ $$ \implies I(x)=\sec(x)\int{\sin(x)}dx-\int{\frac{d(\sec(x))}{dx} ( \int{\sin(x)dx}})dx $$ $$\implies I(x)=\sec(x)(-\cos(x))-\int{\sec(x) \tan(x)(-\cos(x))}dx$$ $$\implies I(x)=-1+I(x)$$

  1. What have I done wrong in the evaluation of the integral? What does it tell about the integral?
  2. If it is correct, are there any limitations to the product rule(integration by parts)? When and how must it be used?
Antiderivatives are unique upto a constant

It is clear from the linked question what my mistake was.
So, I just wanted to know, if a particle had a velocity of $v_1=\tan t $ and another had a velocity of $v_2=\frac{\sin t}{\cos t}$ and both start from rest and from same position(x,y). The constant 'C' is known for $v_1$, then why will the value of displacement be $1 \ more \ than \ that \ of \ v_2 \ (evaluated \ the \ 2^{nd} way) $
Basically why does $s_2=-1+s_1$(where s is displacement and t is time)

This question, the comments and the answer kind of dismiss the ILATE thing as simply a tool for faster simpler calculation so there is no fixed sequence of assuming u and v and that cannot be the mistake.

Aurelius
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  • I've taken the liberty to tidy up the TeX a bit: remember that common math functions render nicely with a backslash. For example \sin x renders $\sin x$ which looks nicer than $sin \ x.$ – Sean Roberson Sep 16 '23 at 18:49
  • Your evaluation is correct! Better questions to ask are “what does your evaluation tell you about the integral?” and “is that information useful?”.
    • – shanksion Sep 16 '23 at 19:06
  • There are limitation to using the product rule. Like how you assumed $u=\sec x$ was differentiable. You may not know, given an arbitrary function, whether this can be assumed or not.

    • When can IBP be used? A: Whenever you satisfy the conditions to apply IBP you can use it. Though it may not always be the most useful tool in the world.

    – shanksion Sep 16 '23 at 19:09
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    Integration by parts should not be called "the product rule." It is not an analogue of the product rule for differentiation; it does not tell you how to integrate products in full generality. It's a trick that simplifies integrals sometimes and often does nothing. – Qiaochu Yuan Sep 16 '23 at 19:17
  • No wrong because antiderivatives are unique . – Delta Sep 16 '23 at 21:29
  • https://math.stackexchange.com/q/2207389/1183190 – Delta Sep 16 '23 at 21:32
  • @Mostafa antiderivative are unique “up to a constant”. – shanksion Sep 16 '23 at 22:05