Evaluate the given integral:

$$I=\int \frac{1}{\sin (x-a) \sin(x-b) \sin(x-c)}.dx$$

When it was $I=\int \frac{1}{\sin (x-a) \sin(x-b)}.dx$, I solved it by multiplying and dividing by $\sin (b-a)$. But I am not getting how to proceed here..

Could someone help me with this?

  • $\begingroup$ It would be helpful to me to know if you really need a primitive or if you are just interested in the integral computed over some interval. A brute-force approach is just to use partial fraction decomposition, since $\frac{1}{\sin(z)}$ is a meromorphic function with known singularities and residues. $\endgroup$ – Jack D'Aurizio Aug 4 '16 at 18:10
  • $\begingroup$ Moreover, we may remove a parameter by a suitable substitution like $x=t+b$, so you problem is equivalent to finding $$\int_{?}\frac{dx}{\sin x}\cdot\frac{1}{\sin(x-u)\sin(x+v)}.$$ $\endgroup$ – Jack D'Aurizio Aug 4 '16 at 18:12

I'd like to expand on @Kajelad explanation. The system can be solved as follows.

We want to perform the decomposition

$$\frac{1}{\sin(x-a)\sin(x-b)\sin(x-c)} = \frac{A}{\sin(x-a)} + \frac{B}{\sin(x-b)} + \frac{C}{\sin(x-c)}$$

or by rewriting we obtain

$$1 = A\sin(x-b)\sin(x-c) + B\sin(x-a)\sin(x-c) + C\sin(x-a)\sin(x-b).$$

As we want this to hold for all $x$ we may select a few values to see if we get a solution. Naturally, the equation should hold if $x=a$ hence we obtain

$$1 = A\sin(a-b)\sin(a-c) + B\sin(0)\sin(a-c) + C\sin(0)\sin(a-b)$$

but as $\sin(0) = 0$ then

$$A = \frac{1}{\sin(a-b)\sin(a-c)}.$$

Similarly, by selecting $x=b$ and $x=c$ we obtain

$$B = \frac{1}{\sin(b-a)\sin(b-c)}$$

$$C = \frac{1}{\sin(c-a)\sin(c-b)}.$$

Through brute work and using the odd property of $\sin$ and trigonometric formulas these should verify that the decomposition is justified. If so, then the solution is given by

$$\int \left(A\csc(x-a) + B\csc(x-b) + C\csc(x-c)\right) dx$$ $$ = \frac{\ln(\tan(x-a))}{\sin(a-b)\sin(a-c)} + \frac{\ln(\tan(x-b))}{\sin(b-a)\sin(b-c)} + \frac{\ln(\tan(x-c))}{\sin(c-a)\sin(c-b)} + C$$


I looked at the partial feaction decomposition method. It is not particularly elegant, but it reduces the calculus problem to a linear algebra one. It is likely a more efficient method for this integral exists, but I am not sure what it would be.



If $A$, $B$, and $C$ are constants, we can use the identity $\int\csc x\ dx=\ln\left(\tan\frac x2\right)+k.$


In order to find these constants, we can rewrite $\sin(x-a)=\sin x\cos a-\sin a\cos x.$

$$=\int\frac{\begin{array}A(\sin x\cos b-\sin b\cos x)(\sin x\cos c-\sin c\cos x)\\+B(\sin x\cos a-\sin a\cos x)(\sin x\cos c-\sin c\cos x)\\+C(\sin x\cos b-\sin b\cos x)(\sin x\cos c-\sin c\cos x)\end{array}}{(\sin x\cos a-\sin a\cos x)(\sin x\cos b-\sin b\cos x)(\sin x\cos c-\sin c\cos x)}dx$$

Rearranging, and rewiring some sines using the sum identity:

$$=\int\frac{\begin{array}(A\cos b\cos c+B\cos a\cos c+C\cos a\cos\ b)\sin^2x\\-(A\sin(b+c)+B\sin(a+c)+C\sin(a+b))\sin x\cos x\\+(A\sin b\sin c+B\sin a\sin c+C\sin a\sin b)\cos^2x\end{array}}{(\sin x\cos a-\sin a\cos x)(\sin x\cos b-\sin b\cos x)(\sin x\cos c-\sin c\cos x)}dx$$

The double angle identity can be used to show that $$\alpha\sin^2x+\beta\sin x\cos x+\gamma\cos^2x=\frac12(\alpha+\gamma)+\frac12(-\alpha+\gamma)\cos 2x+\beta\sin 2x,$$ which is equal to $1$ for all $x$ if and only if $\alpha=1,\beta=0,\gamma=1$. This gives a linear system of equations.

$$A\cos b\cos c+B\cos a\cos c+C\cos a\cos b=1$$ $$A\sin b\sin c+B\sin a\sin c+C\sin a\sin b=1$$ $$A\sin(b+c)+B\sin(a+c)+C\sin(a+b)=0$$

Rewriting as a matrix:

$$\begin{bmatrix}\cos b\cos c & \cos a\cos c & \cos a\cos b\\ \sin b\sin c & \sin a\sin c & \sin a\sin b\\ \sin(b+c) & \sin(a+c) & \sin(a+b)\\\end{bmatrix}\begin{bmatrix}A\\B\\C\\\end{bmatrix}=\begin{bmatrix}1\\1\\0\\\end{bmatrix}$$

For specific values of $a$, $b$, and $c$, this equation above can be solved by row reduction. Technically, the inverse of the $3\times3$ matrix above can be computed to find the general solution, but it would be a rather cumbersome process.


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