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Is there a nicer way to compute this integral?? What's the trick to this? Here's my attempt:

$$\int_{0}^{\pi}\sin(x)\sqrt{\frac{1+\cos(9x)}{1+\cos(x)}}dx=\int_{0}^{\pi} \tan(x)\cos(\frac{9x}{2})dx=\frac{1}{2}\int_{0}^{\pi}\frac{\sin(\frac{11x}{2})-\sin(\frac{7x}{2})}{\cos(x)}dx$$

As you can see, this is not a pretty integral.

Silver
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  • try https://math.stackexchange.com/questions/439851/evaluate-the-integral-int-frac-pi2-0-frac-sin3x-sin3x-cos3x/439856#439856 – lab bhattacharjee Nov 16 '23 at 12:30
  • Do you know this trick: $1+\cos{x} = 2 \cos^2({\frac{x}{2}})$, this can be very useful while handling square roots of $1+\cos{x}$. – Dominique Nov 16 '23 at 13:34

2 Answers2

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Using double angle identities the integrand simplifies to

$$\int_0^\pi 2\sin\frac{x}{2}\left|\cos\frac{9x}{2}\right|dx = \int_0^\pi |\sin 5x - \sin 4x |dx$$

The absolute value in the denominator can be ignored when simplifying since it is always nonnegative on the specified domain. However this integral's answer is the messy

$$\frac{1}{20}\left(17+18\cos\frac{\pi}{18}+18\cos\frac{\pi}{9}+18\cos\frac{2\pi}{9}\right)$$

after taking into account the zero crossings of the integrand. The answer without the absolute value would have been $\frac{1}{5}$, and maybe that was a mistake on the problem writer's part.

Ninad Munshi
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I started an answer... and in between we have already an accepted one. Well, the result slightly differs, so i completed the message, and will post it...

Let us denote by $J$ the integral to be computed, we have: $$ \begin{aligned} J &= \int_0^\pi \sin x\sqrt{\frac{1+\cos(9x)}{1+\cos x}}\; dx \qquad\text{(Substitution: $x=2y$)} \\ & = \int_0^{\pi/2} \sin (2y)\sqrt{\frac{1+\cos(18y)}{1+\cos (2y)}}\; 2\; dy = \int_0^{\pi/2} \sin (2y)\sqrt{\frac{2\cos^2(9y)}{2\cos^2 y}}\; 2\; dy \\ &= 2\int_0^{\pi/2} \sin (2y)\frac{|\ \cos(9y)\ |}{\cos y}\; dy \qquad\text{(Substitution: $y=\frac \pi2-u$)} \\ & = 2\int_0^{\pi/2} \sin (2u)\frac{|\ \sin(9u)\ |}{\sin u}\; du = 2\int_0^{\pi/2} 2\cos u\;|\ \sin(9u)\ |\; du \\ &= \left( \int_0^{\pi/9} - \int_{\pi/9}^{2\pi/9} + \int_{2\pi/9}^{3\pi/9} - \int_{3\pi/9}^{4\pi/9} + \int_{4\pi/9}^{\pi/2} \right) \left(-\frac 14\cos(8u)-\frac 15\cos(10u)\right)'\; du \\ &= \frac {17}{20} +\frac 9{10} \underbrace{ \left( \cos\frac \pi9 + \cos\frac {2\pi}9 + \cos\frac {4\pi}9 \right)}_{:=g} \approx 2.541446717414\dots\qquad\ . \end{aligned} $$ Note: We can compute the "Gauss sum" denoted above by $g$ "(more) explicitly", $$ \begin{aligned} g &= \cos\frac \pi9 + \cos\frac {2\pi}9 + \cos\frac {4\pi}9 = \cos\frac \pi9 + \cos\frac {(3-1)\pi}9 + \cos\frac {(3+1)\pi}9 \\ &= \cos\frac \pi9 + 2\cos \frac {3\pi}9\cos\frac \pi 9 =2\cos\frac \pi 9 \end{aligned} $$ So we finally have: $$ \bbox[lightyellow]{\qquad J=\frac {17}{20}+\frac 95\cos\frac \pi9\ .\qquad} $$

Numerical checks, here pari/gp:

If we use a high precision (1000) in pari, then

intnum( x=0, Pi, sin(x) * sqrt( (1+cos(9*x)) / (1+cos(x)) ) )

delivers 2.5414565207310... and the computation of the above claimed value is

? \p 60
   realprecision = 77 significant digits (60 digits displayed)
? 17/20 + 9/5*cos(Pi/9)
%42 = 2.54144671741463509129739669918451664588517464167603633956252

Well, the numeric integration delivers at least four exact decimals.

dan_fulea
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