special technique on definite integral

Posted: October 24, 2012 in Mathematics
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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,

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

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.

\displaystyle\int\frac{dx}{(x^3+1)(x^2+1)}

Partial fractions yields,

\displaystyle=\int-\frac{2x-1}{3(x^2-x+1)}+\frac{x+1}{2(x^2+1)}+\frac{1}{6(x+1)}\;dx

\displaystyle=-\int\frac{d(x^2-x+1)}{3(x^2-x+1)}+\int\frac{d(x^2+1)}{4(x^2+1)}+\int\frac{dx}{2(x^2+1)}+\int\frac{d(x+1)}{6(x+1)}

\displaystyle=-\frac{\log(x^2-x+1)}{3}+\frac{\log(x^2+1)}{4}+\frac{\tan^{-1}x}{2}+\frac{\log|x+1|}{6}+C

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}

Therefore,

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

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