I am away the integration by part for a long time .
I don't remember if the necessary condition for integrating by part .
I saw this $Wrong \space solution$ but I can't find "What is behind ...?"
This question is about Evaluating ($\text{Bloom's taxonomy}$).
$$\int \cot xdx=\int \frac{\cos x}{\sin x}dx=$$ take integration by part as below
$$du=\cos x dx \to u=sin x\\v=\frac{1}{\sin x} \\
\to \int \frac{\cos x}{\sin x}dx=\sin x \cdot \frac{1}{\sin x}-\int \sin x\cdot\frac{0-\cos x}{\sin^2 x}dx=\\1-(-)\int \frac{\cos x}{\sin x}dx=\\1+\int \frac{\cos x}{\sin x}dx=\\\int \cot xdx=
$$
Can someone help me "What's wrong going here" ?
Thanks in advance .
It should become clear what is going on here when you write the integration by parts formula for definite integrals ...
$$ \int_a^bu\;dv = uv|_a^b - \int_a^bv\;du$$
In your case you have $uv=1$ so $ uv|_a^b= 0$ your result comes done to the tautology
$$ \int_a^b\cot x \; dx = \int_a^b\cot x \; dx $$
which is not wrong, but neither is it productive.
This integral can be determined using straightforward substitution $u=\sin x$, yielding the result
$$ \int_a^b\cot x \; dx = \ln|\sin x| \; \; |_a^b = \ln \left| \dfrac {\sin b}{\sin a} \right| $$
Perhaps there is also an understanding issue here about the fundamental difference between applying techniques of integration and techniques of integration.
In the case of differentiation any applicable rule will necessarily produce the derivative (provided that you can differentiate the constituent functions.) If more than one technique can be applied, they must all produce the same result.
With integration there is no guarantee that any technique will produce a closed form result ( e.g. $\int e^{-x^2}dx$). If one technique does not produce a result, it is quite possible that another technique will prove fruitful.
I think this is related to the fact that the power rule $$(fg)' = f'g + fg'$$ includes only the derivatives of the individual functions $f$ and $g$
By contrast, in the integration by parts formula $$ \int_a^b(fg')\;dx = (fg)|_a^b - \int_a^b(f'g)\;dx$$ the right side contains the integral of a product.