What can using the "opposite" combination for integration by parts be used for For example, 
$$I=\int xe^{x}dx$$
By taking the derivative of x, and then repeating integration by parts once, the integral can be evaluated trivially. However, when taking the derivative of $e^{x}$ and integrating $x^2$, the process goes on forever. Another example would be 
$$J=\int \sin{x}\cos{x}dx$$
By repeating integration by parts, 
$$J= sin^2x + cos^2x + sin^2x + cos^2x ....$$
Does this have any use/any interesting links? e.g. could be used to evaluate integrals where other "combination" of choosing which term is differentiated/integrated is not plausible. 
 A: To expand on my comment, I will explicitly use your first example and another example I have.
For $I(z) = \int_0^z xe^x\:dx$, going in the "opposite" direction gives us
$$I(z) = \frac{1}{2}z^2e^z - \int_0^z \frac{1}{2}x^2e^x\:dx \implies I(z) = e^z\left(\frac{1}{2}z^2-\frac{1}{6}z^3+\cdots\right)$$
$$=e^z\left(e^{-z}-1+z\right)=1-e^z+ze^z$$
getting the series from repeated integration by parts. Another example would be
$$\int_0^{\frac{1}{2}}\frac{1}{\sqrt{1-x^2}}\:dx = \frac{x}{\sqrt{1-x^2}}\Biggr|_0^{\frac{1}{2}}-\int_0^{\frac{1}{2}}\frac{x^2}{(1-x^2)^{\frac{3}{2}}}\:dx$$
which gives a series for
$$\arcsin\left(\frac{1}{2}\right) = \frac{\pi}{6} = \frac{1}{\sqrt{3}}-\frac{1}{3}\left(\frac{1}{\sqrt{3}}\right)^3+\frac{1}{5}\left(\frac{1}{\sqrt{3}}\right)^5-\cdots$$
A: On example I could see is the following :
Imagine you want to compute
$$\int_{0}^{1}\frac{\text{d}x}{\left(1+x^2\right)^2}$$
You can use that
$$
\int_{0}^{1}\frac{\text{d}x}{1+x^2}=\frac{\pi}{4}
$$
and then you integrate by part "in the other way" you usually do
$$
\int_{0}^{1}\frac{\text{d}x}{1+x^2}=\int_{0}^{1}\frac{1}{1+x^2}\times 1\text{ d}x=\left[\frac{x}{\left(1+x^2\right)^2}\right]^{1}_0+\int_{0}^{1}\frac{2x^2}{\left(1+x^2\right)^2}\text{d}x
$$
Hence
$$ \frac{\pi}{4}=\frac{1}{4}+2\int_{0}^{1}\left(\frac{1}{1+x^2}-\frac{1}{\left(1+x^2\right)^2}\right)\text{d}x$$and we find

$$\int_{0}^{1}\frac{\text{d}x}{\left(1+x^2\right)^2}=\frac{2+\pi}{8}$$

