Can $\int \frac{\sec x \ \mathrm{d}x}{\sqrt{\sin(x+2A)+\sin(A)}}$ be evaluated using elementary functions?

This question recently popped up on one of my tests and I still have no idea on how to even begin.

I tried assuming $$\sin(x+2A)+\sin(A)=t^2$$ which did not help at all.

Then I went on to multiply and divide by $$\tan x$$ to create a derivative on the numerator, still does not help.

I even tried to expand $$\sin(x+2A)$$ which too does not seem to follow.

I have also tried Approach$$0$$ and Wolfram Alpha but they do not seem to help.

and many more...

Now I am stumped I have no idea as to how to even begin. I have tried everything that I could think of.

Note that we are only familiar with basics of integration(no contour integrals ; no special functions etc..)

$$\bullet~\textbf{Question:}~$$

"Can the above integral be evaluated in elementary functions?" If yes then how? If not , can we prove that it cannot be solved using elementary funcions?

• Are you sure this is not a definite integral from, say $-A$ to $A$? Commented Jul 20, 2020 at 9:04
• @AniruddhaDeb My test had only this indefinite integral. I am not sure if it could be a typo as the answers are not yet released (and not likely to be) Commented Jul 20, 2020 at 9:06
• The expression in the denominator looks like it could be reduced with the property $\int_a^b f(x) = \int_a^b f(a+b-x)$, which is why I asked. Commented Jul 20, 2020 at 9:08
• Commented Jul 20, 2020 at 9:17
• Commented Jul 20, 2020 at 9:44

2 Answers

Not elementary at all

$$I=\int \frac{\sec (x)}{\sqrt{\sin(x+2a)+\sin(a)}}\,dx$$ $$\sin(x+2a)+\sin(a)=\sin (a)+\sin (2 a) \cos (x)+\cos (2 a) \sin (x)$$ Using the tangent half-angle substitution $$x=2 \tan^{-1}(t)$$ $$\sin(x+2a)+\sin(a)=\frac{(\sin (a)+\sin (2 a))+2 t \cos (2 a)+t^2 (\sin (a)-\sin (2 a))}{1+t^2}$$ $$(\sin (a)+\sin (2 a))+2 t \cos (2 a)+t^2 (\sin (a)-\sin (2 a))=$$ $$(\sin (a)-\sin (2 a))\left(t+\tan \left(\frac{3 a}{2}\right)\right) \left(t-\cot \left(\frac{a}{2}\right)\right)$$ Let $$\alpha=\sin (a)+\sin (2 a)\qquad \beta=\tan \left(\frac{3 a}{2}\right) \qquad \gamma=\cot \left(\frac{a}{2}\right)$$

$$I=\frac 2 \alpha \int \frac {\sqrt{1+t^2}} {(1-t^2)\sqrt{(t+\beta) (t-\gamma) }}\,dt$$

The result is given in terms of elliptic integrals (have a look here).

Edit

$$a=\frac \pi 4 \qquad \qquad \implies \qquad I=2 \sqrt{2-\sqrt{2}} \,\,\Pi \left(2;\frac{x}{2}|4-2 \sqrt{2}\right)$$

Let $$A=\dfrac\pi2-2b,\quad x=4b-\dfrac\pi2-y,\tag1$$ then $$I=\int\dfrac{\text dy}{\cos(y-4b)\sqrt{\cos y+\cos2b}} =\int\dfrac{\text dy}{(\cos4b\cos y+\sin4b\sin y)\sqrt{\cos y+\cos2b}}$$ $$=\int\dfrac{(\cos4b\cos y-\sin4b\sin y)\,\text dy}{(\cos^2 4b\cos^2 y-\sin^2 4b\sin^2y)\sqrt{\cos y+\cos2b}}=I_1\cos4b-I_2\sin4b,$$ where $$I_1=\int\dfrac{\cos y\,\text dy}{(\cos^2y-\sin^24b)\sqrt{\cos y+\cos2b}}, \tag2$$ $$I_2=\int\dfrac{\sin y\,\text dy}{(\cos^2y-\sin^2 4b) )\sqrt{\cos y+\cos2b}}.\tag3$$ Integral $$\,I_2\,$$ can be expressed via elementary functions (see the WA result below).

At the same time, the first integral in the common case can be expressed via elliptic integral of the third kind and cannot be expressed via elementary functions.

However, for $$\;b=\dfrac\pi2 k,\;A=\dfrac\pi2+\pi n,\; k,n\in\mathbb Z\;$$ there are exceptions