I recently stumbled upon an indefinite integral . sin(x)/x [ Another similar one is root (x) times sin x.
However if we substitute sin(x) in terms of x as Maclaurin series we could get a series of infinite yet integrable polynomials . What's the catch?
Answer
The quote you're referring to is "It is worth mentioning that integration by parts is not applicable to product of functions in all cases. For instance, the method does not work for $\int \sqrt{x} \sin\; x \; dx$.
The reason is that there does not exist any function whose derivative is
$\sqrt{x} \sin \; x$."
As stated, this is simply wrong. A correct statement would be that there is not an elementary function whose derivative is $\sqrt{x} \sin\; x$. Integration by parts "works" on $\sqrt{x} \sin\; x$, but it doesn't give you a closed form for the antiderivative. It will say, e.g., that
$$ \int \sqrt{x} \sin \; x \; dx = \frac{2}{3} x^{3/2} \sin\; x - \frac{2}{3} \int x^{3/2} \cos\; x\; dx $$
which is correct, but not really helpful: it leaves you no closer to finding a formula than before.
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