TITLE: Two interesting results in integration $\int_{0}^{a}f(a-x) \ \mathrm{d}x= \int_{0}^{a}f(x)\ \mathrm{d}x$ and differentiation of powers of functions
QUESTION [6 upvotes]: I am investigating the following result in integration
$\displaystyle\int_{0}^{a}f(a-x) \ \mathrm{d}x = \int_{0}^{a}f(x) \ \mathrm{d}x \ \ \ (*)$
This neat little result forms the basis for many questions in calculus exams, often then asking one to evaluate something like
$\displaystyle\int_{0}^{\frac{\pi}{2}}\frac{\sin^n x}{\sin^n x + \cos^n x} \ \mathrm{d}x$ where $n$ is a positive integer. The process of solving this integral isn't too challenging, and is almost immediate from $(*)$. 
My question is this: can anyone think of any more challenging integrals out there (possibly requiring some clever substitution, integration by parts etc.) that $(*)$ can help solve?
UPDATE
I also came across another identity involving differentiation: 
$\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}(u(x))^{v(x)} = (u(x))^{v(x)}\left(\frac{\mathrm{d}v(x)}{\mathrm{d}x}\ln u(x) + \frac{v(x)}{u(x)}\frac{\mathrm{d}u(x)}{\mathrm{d}x}\right)$. 
This is another identity that can be used to solve integrals, but I am again unable to find any creative examples, so if anyone could suggest some I'd be happy to give them a go.

REPLY [2 votes]: There are a lot of possible answers. For example, $$\int_{0}^{1} \frac{x^3}{3x^2-3x+1} \mathrm{d} x=\int_{0}^{1} \frac{x^3}{x^3+(1-x)^3} \mathrm{d} x=\frac{1}{2}$$ or $$\int_{0}^{1}\frac{x^5}{5x^4-10x^3+10x^2-5x+1}\mathrm{d}x=\frac{1}{2}$$ are both good examples of how this property can be used. We can use this property to calculate these complicated looking integrals in less than a few seconds. 
If we were not to use this property, we would have to use things like $$\int \frac{x^5}{5x^4-10x^3+10x^2-5x+1}\mathrm{d} x$$
Which ends up being more than slightly complicated, as can be seen here. .
In general, we have the property

$$\int_{0}^{1} \frac{x^{2n+1}}{\sum_{k=1}^{2n+1}\binom{2n+1}{k} (-x)^{2n+1-k}}\mathrm{d}x=\frac{1}{2}$$