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The integral is $$\int \frac{\mathrm{d}x}{\sin^2 x+\sin x+1}$$

Consider $$\int \frac{\mathrm{d}x}{ax^2+bx+c}$$ There are a total of three cases, depending on the discriminant of $ax^2+bx+c$. Two of which are shown here (#1).

The third one, i.e. when $\Delta=0$, simply means evaluating $$\frac{1}{a}\int \frac{\mathrm{d}x}{\big(x+\frac{b}{2a}\big)^2} $$

For the first two cases, you can see different substitutions are used so as to prevent $i$ from appearing in the answer, hence resulting in an $\arctan$ function and a $\ln$ function respectively. The discrepancy arises when we change $+(4ac-b^2)$ to $-(b^2-4ac)$.

In general, is it possible to evaluate $$\int \frac{\mathrm{d}x}{a\sin^2 x+b\sin x+c}$$ such that the result does not contain complex numbers when $\Delta_{\sin x}<0$?

The only approach I can think of when $\Delta_{\sin x}>0$ is by partial fraction decomposition, which differs from the method of substitution used above.

L Parker
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3 Answers3

4

Let $\displaystyle I = \int \frac {1}{\sin ^2x + \sin x + 1}dx$

Let $$\sin x = -\frac {(2-\sqrt {3})t + (2 +\sqrt {3})}{t + 1}\implies \cos x\ dx = \frac {2\sqrt {3}}{(t + 1)^2}dt$$

Then $$I = 2\sqrt 3 \int \frac {t + 1}{\left((6 - 3\sqrt 3)t^2 + (6 + 3\sqrt 3)\right)\sqrt {(4\sqrt {3} - 6)t^2 - (4\sqrt {3} + 6)}}\ dx$$

Now consider $$I_1 = \int \frac {t}{\left((6 - 3\sqrt 3)t^2 + (6 + 3\sqrt 3)\right)\sqrt {(4\sqrt {3} - 6)t^2 - (4\sqrt {3} + 6)}}\ dx$$

and substitute $ \ (4\sqrt {3} - 6)t^2 - (4\sqrt {3} + 6) = z^2$ Reduce into well known equation.

and Consider $$ \ I_2 = \int \frac {1}{\left((6 - 3\sqrt 3)t^2 + (6 + 3\sqrt 3)\right)\sqrt {(4\sqrt {3} - 6)t^2 - (4\sqrt {3} + 6)}}\ dx$$

and first put $\displaystyle \ t = \frac 1p$ after this substitution put $ \ (4\sqrt {3} - 6) - (4\sqrt {3} + 6)p^2 = u^2$

and we are done without using Complex number...

juantheron
  • 56,203
2

Let $\sin x =\frac{1-t^2}{1+t^2}$, or $t= \tan(\frac\pi4-\frac x2)$ \begin{align} &\int \frac{1}{\sin^2 x+\sin x+1}\ dx =\int\frac{2+2{t^2}}{3+t^4} \ dt\\ =&\ (1+\frac1{\sqrt3}) \int \frac{1+\frac{\sqrt3}{t^2}}{t^2+\frac3{t^2}}dt +(1-\frac1{\sqrt3}) \int \frac{1-\frac{\sqrt3}{t^2}}{t^2+\frac3{t^2}}dt\\ =&\ \frac{\sqrt3+1}{\sqrt{6\sqrt3}}\tan^{-1}\frac{t-\frac{\sqrt3}t}{\sqrt{2\sqrt3}} -\frac{\sqrt3-1}{\sqrt{6\sqrt3}}\coth^{-1}\frac{t+\frac{\sqrt3}t}{\sqrt{2\sqrt3}}+C \end{align}

Quanto
  • 120,125
1

The answer is yes.

Simply divide throughout by either $\sin^2{x}$ or $\cos^2{x}$

and respectively substitute $c = \cot{x}$ or $t = \tan{x}$

Jack Tiger Lam
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