I can integrate $\sec^a\theta$, where $a $ is even, without any difficulty. It is simply the integral of the binomial expansion of $(u^2+1)^{\frac{a-2}{2}}$. (Where $u=\tan\theta$)

When $a$ is odd, things get trickier. I have to integrate the even function ( integral of the binomial expansion of $(u^2+1)^{\frac{a-3}{2}},\space where \space u=\tan\theta,$) and the residual $\sec\theta$ by parts. But this is messy.

So my question is: Does the binomial theorem hold for fractional exponents so that $$\int\sec^a\theta= \int \text{the binomial expansion of }(u^2+1)^{\frac{a-2}{2}},\space \text{where} \space u=\tan\theta,$$ for ALL rational values of "$a$".

  • $\begingroup$ I could barely imagine that it would hold if $a$ is irrational:) $\endgroup$ – imranfat Feb 14 '14 at 15:58
  • $\begingroup$ True, edited to clarify. $\endgroup$ – Chris Feb 14 '14 at 16:05
  • $\begingroup$ If a is irrational and undefined boundaries, you will need an hypergeometric function. $\endgroup$ – JJacquelin Feb 14 '14 at 16:26

Yes, the binomial theorem holds for noninteger exponents; this is a famous discovery of Newton's. See binomial series.

And, by the way, here's another method for the case of odd integers: $$ \int \sec^{2k+1}\theta \,d\theta = \int \frac1{\cos^{2k+2}\theta} \cos\theta\,d\theta = \int \frac1{(1-u^2)^{k+1}} \,du $$ and now deploy partial fractions. (According to Eli Maor's book Trigonometric Delights, this is actually how Isaac Barrow first computed $\int\sec\theta\,d\theta$, which is notable as the first known use of partial fractions for integration.)

  • $\begingroup$ Just wondering, Is the unpaired 2 in the second expression supposed to be a 1? $\endgroup$ – Chris Feb 14 '14 at 17:20
  • $\begingroup$ No, that extra $\cos\theta$ is to cancel the $\cos\theta$ introduced up top. $\endgroup$ – user21467 Feb 14 '14 at 17:36
  • $\begingroup$ Of course, don't know how I missed it. $\endgroup$ – Chris Feb 14 '14 at 17:43

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