# Difficult Definite Integral $\int_{0}^{\frac{\pi}{2}} \sqrt{1+2\cos^2\left(\frac{\pi}{2} - x\right)} + \sin x\, dx$

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:

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

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

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

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

The integral then became:

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

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.

• see wolframalpha.com/input/… do you have any typo? Dec 21, 2018 at 1:45
• I second @MartínVacasVignolo's comment; as written you can't really get a nice closed for answer for this. Dec 21, 2018 at 1:47
• 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. Dec 21, 2018 at 1:47
• @austintice Then I would email them to get clarification if this is a homework problem and has a due date. Dec 21, 2018 at 1:48
• @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. Dec 21, 2018 at 1:50

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 $$E(-2)\approx 2.184438$$.

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.