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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+2cos^2left(frac{pi}{2} - xright)} + sin x, dx
end{equation}

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

The integral then became:
begin{equation}
int_{0}^{frac{pi}{2}} sqrt{1+2sin^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.

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

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

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

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 $0leq k leq 4$.

Using $k=-2$, we should get , as an approximation, $frac{5810969}{8357946}piapprox 2.18423$ 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, $frac{214931493555 }{309110015222}piapprox 2.184424$ while the exact value would be $E(-2)approx 2.184438$.