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?


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.

  • $\begingroup$ math.stackexchange.com/questions/1595361/… $\endgroup$
    – user160738
    Feb 8, 2017 at 21:30
  • $\begingroup$ That's certainly an interesting one, but there's not much more to it than the example I gave. I was looking to see if it can be used to simplify more difficult integrals, where the method is less immediately obvious. $\endgroup$
    – wrb98
    Feb 8, 2017 at 21:33
  • $\begingroup$ Another one I found was $\displaystyle \int_{3}^{9}\frac{\ln|x-9|}{\ln|x-3|+\ln|x-9|}\ \mathrm{d}x$, but again it is immediately obvious how to proceed. $\endgroup$
    – wrb98
    Feb 8, 2017 at 21:41
  • $\begingroup$ Try $\int_0^\frac{\pi}{2}\ln (\sin x) dx$ with that method. $\endgroup$
    – Shashi
    Feb 9, 2017 at 13:30
  • $\begingroup$ @Shashi That's a pretty neat one too! There are some really fun examples out there. Could you suggest an integral that my other identity could help to solve? (I updated the question) This other identity is less well-known, and I cannot find that many nice examples. $\endgroup$
    – wrb98
    Feb 11, 2017 at 20:34

2 Answers 2


Try this one! $$\int_{0}^{1}\frac{\ln(1+x)}{1+x^{2}}dx$$

  • $\begingroup$ This is a good one! I also came across another very nice identity, which I have added to my original post. Can you think of any interesting integrals that this new identity can help solve? $\endgroup$
    – wrb98
    Feb 11, 2017 at 20:24
  • $\begingroup$ Since neither answer addressed the second identity in the update, I shall have to award the 50 point bounty to the answer which I thought gave the most interesting example of where the first identity is useful, which in this case was courtesy of BaroqueFreak $\endgroup$
    – wrb98
    Feb 22, 2017 at 14:29

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

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

  • $\begingroup$ Why the downvote? $\endgroup$
    – S.C.B.
    Feb 20, 2017 at 0:23
  • $\begingroup$ i upvoted you ! $\endgroup$
    – satyatech
    Feb 21, 2017 at 12:48
  • $\begingroup$ That's a good one for the first identity. Can you think of any integrals that are made easier by the second one? $\endgroup$
    – wrb98
    Feb 22, 2017 at 11:36
  • $\begingroup$ @wrb98 What is the first identity? $\endgroup$
    – S.C.B.
    Feb 22, 2017 at 11:37
  • $\begingroup$ The one involving substitution with $f(a-x)$, the second one I'm referring to is the one in the update (the differentiation of powers) $\endgroup$
    – wrb98
    Feb 22, 2017 at 12:51

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