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I have spent several days trying to solve this integral, but to no avail. This isn't from a textbook, but a challenge problem given to me by a professor. I am not looking for anyone to give me the solution, but just to lead me in the right direction.

The problem is to compute the following integral:

\begin{equation} \int_{0}^{\frac{\pi}{2}} \sqrt{1+2\cos^2\left(\frac{\pi}{2} - x\right)} + \sin x\, dx \end{equation}

When first approaching this problem I tried to utilize the cofunction identity: \begin{equation} \cos\left(\frac{\pi}{2}-x\right) = \sin x \end{equation}

The integral then became: \begin{equation} \int_{0}^{\frac{\pi}{2}} \sqrt{1+2\sin^2x} + \sin x\, dx \end{equation}

I have tried several things from this point such as using the formulas \begin{equation} \sin^2x = \frac{1}{2}[1-\cos(2x)] \end{equation}

The integral then became:

\begin{equation} \int_{0}^{\frac{\pi}{2}} \sqrt{2-\cos(2x)} + \sin x\, dx \end{equation}

The issue is I have tried several run arounds(of which I will not post each) with identities and other methods, but I seem to be hitting dead ends. Also, I want to mention that I'm trying to solve this using elementary methods only. I only have experience up to calculus II. Any constructive criticism or comments would be greatly appreciated! Thank you.

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  • $\begingroup$ see wolframalpha.com/input/… do you have any typo? $\endgroup$ Dec 21, 2018 at 1:45
  • $\begingroup$ I second @MartínVacasVignolo's comment; as written you can't really get a nice closed for answer for this. $\endgroup$ Dec 21, 2018 at 1:47
  • $\begingroup$ No typo as far as I'm aware. The person who posed the problem may have meant to put something else, but this is the integral I was given. $\endgroup$
    – rooted
    Dec 21, 2018 at 1:47
  • $\begingroup$ @austintice Then I would email them to get clarification if this is a homework problem and has a due date. $\endgroup$ Dec 21, 2018 at 1:48
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    $\begingroup$ @DeficientMathDude like I stated this isn't a homework problem, but merely a challenge integral problem given to me by one of my research professors to take a shot at over winter break. I have given this problem ~week of time, but just keep getting looped around. I did look up and see the elliptic integral on wikipedia, but was not aware of this before. $\endgroup$
    – rooted
    Dec 21, 2018 at 1:50

2 Answers 2

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As said in comments and answers, you are facing elliptic integrals that you cannot evaluate easily. $$\int_0^{\frac \pi 2}\sqrt{1+k \sin ^2(x)}\,dx=E(-k)$$ where appears the complete elliptic integral of the second kind.

However, we can build quite good approximations. I give you one I produced years ago (for rather small values of $k$) using Padé approximants built at $k=0$.

$$E(-k) \simeq \frac \pi 2 \,\frac{1+\frac{39575 }{28464}k+\frac{20621} {37952}k^2+\frac{129235}{2428928}k^3 } {1+\frac{32459}{28464}k+\frac{34741 }{113856}k^2+\frac{79037 }{7286784}k ^3 }$$ which is quite good for the range $0\leq k \leq 4$.

Using $k=-2$, we should get , as an approximation, $\frac{5810969}{8357946}\pi\approx 2.18423$ while the exact value would be $E(-2)\approx 2.18444$.

Edit

The approximation I wrote was made more than fourty years ago and it was, at that time, a hard work. Just for the fun of it, I made, after answering, a better one which took me a few minutes .... thanks to a CAS. It is $$E(-k) \simeq \frac \pi 2 \,\frac{1+\frac{133542997 }{70902928}k+\frac{1325913585 }{1134446848}k^2+\frac{1210596065 }{4537787392}k^3+\frac{4808786003 }{290418393088} k^4} {1+\frac{115817265 }{70902928}k+\frac{915821721 }{1134446848}k^2+\frac{553597479 }{4537787392}k^3+\frac{777708891 }{290418393088}k^4 }$$ Using $k=-2$, we should get , as an approximation, $\frac{214931493555 }{309110015222}\pi\approx 2.184424$ while the exact value would be $E(-2)\approx 2.184438$.

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You cannot solve the integral using elementary methods. It can be written in terms of a special function called the elliptic integral of the second kind $E(m)$, defined as

$$ E(m) = \int_{0}^{\pi/2}\sqrt{1-m\sin^{2}x}\,\mathrm{d}x. $$

This function has a power series, but that series is also hard to derive without using other special functions.

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