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)$$
- What have I done wrong in the evaluation of the integral? What does it tell about the integral?
- 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.