special technique on definite integral

Posted: October 24, 2012 in Mathematics

As we all know integration doesn’t always have a closed form solution therefore it is harder than differentiation in general.  So, solving an integration problem sometimes require a bit of creativity and luck.  Last week I saw a post on a forum.  The question is to evaluate the following definite integral,


What grabs my attention is the guy who posts this said his teacher can’t do it, then not long after someone replies the poster a solution with closed form answer for the indefinite integral with partial fraction.  But he didn’t bother to plug the upper and lower limits back in.  I guess it’s too tedious to do.


Partial fractions yields,




Then plug in the upper limit and lower limit and after 5 minutes of boring calculation, the answer would be \displaystyle\frac{\pi}{12}

This whole process can take up 15 minutes to an hour to do depending on a person’s math skill.  If you don’t believe me, you can try it on a piece of paper.

But there is a good news, you can solve this in a minute!  This technique isn’t often taught in many schools.  Maybe because not many people have known it or seen it before.  Anyway here it is,

Let \displaystyle x=\frac{1}{w}, hence \displaystyle dx=-\frac{1}{w^2}dw

Then we have,

\displaystyle I=\int_{1/\sqrt{3}}^{\sqrt{3}}\frac{dx}{(x^3+1)(x^2+1)}=\int_{1/\sqrt{3}}^{\sqrt{3}}\frac{w^3\;dw}{(w^3+1)(w^2+1)}

Since w is just a dummy variable in the definite integral.  We have,

\displaystyle I=\int_{1/\sqrt{3}}^{\sqrt{3}}\frac{dx}{(x^3+1)(x^2+1)}=\int_{1/\sqrt{3}}^{\sqrt{3}}\frac{x^3\;dx}{(x^3+1)(x^2+1)}

Now we can add them together as following,

\displaystyle 2I=\int_{1/\sqrt{3}}^{\sqrt{3}}\frac{dx}{(x^2+1)}=\tan^{-1}x\Bigg|_{1/\sqrt{3}}^{\sqrt{3}}=\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}


\displaystyle I=\frac{\pi}{12}


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