How to integrate: $\int \frac{\mathrm dx}{\sin (x)-\sin(a)}$

Question

How to integrate :

$$\int \frac{\mathrm dx}{\sin (x)-\sin(a)}$$

Answer 1

Using Weierstrass substitution, $$\tan \frac x2=u$$

$$\implies \sin x=\frac{2u}{1+u^2}\text{ and } x=2\arctan u,dx=\frac{2du}{1+u^2}$$

$$I=\int\frac{dx}{\sin x-\sin \alpha} =\int\frac1{\frac{2u}{1+u^2}-\sin\alpha}\cdot\frac{2du}{1+u^2} =\int\frac{2du}{2u-(1+u^2)\sin\alpha}$$

Now, $$2u-(1+u^2)\sin\alpha=-\sin\alpha (1+u^2-2u\csc\alpha)=-\sin\alpha \left((u-\csc\alpha)^2-(\cot\alpha)^2\right)$$

Using $\frac{dx}{x^2-a^2}=\frac1{2a}\ln \left|\frac{x-a}{x+a}\right|+C$

$$I=-\frac1{\sin\alpha}\frac1{\cot\alpha}\ln\left|\frac{u-\csc\alpha-\cot\alpha}{u-\csc\alpha+\cot\alpha}\right|+C$$ where $C$ is an arbitrary constant for indefinite integral

Using $\csc\alpha+\cot\alpha=\frac{1+\cos\alpha}{\sin\alpha}=\frac{2\cos^\frac\alpha2}{2\sin\frac\alpha2\cos\frac\alpha2}=\cot\frac\alpha2$ and similarly, $\csc\alpha-\cot\alpha=\tan\frac\alpha2$ (as $\sin2A=2\sin A\cos A,\cos2A=2\cos^2A-1$)

$$I=-\frac1{\cos\alpha}\ln\left|\frac{\tan\frac x2-\cot\frac \alpha2}{\tan\frac x2-\tan\frac\alpha2}\right|+C$$

Again, $$\ln\left|\frac{\tan\frac x2-\cot\frac \alpha2}{\tan\frac x2-\tan\frac\alpha2}\right|$$

$$=\ln\left|\frac{\cos\frac \alpha2\cos\frac x2\left(\sin\frac x2\sin \frac \alpha2-\cos\frac \alpha2\cos\frac x2\right)}{\sin\frac \alpha2\cos\frac x2\left(\sin\frac x2\cos\frac\alpha2-\sin\frac\alpha2\cos\frac x2\right)}\right|=\ln\left|-\cot\frac\alpha2\right|+\ln\left|\frac{\cos\frac{x+\alpha}2}{\sin\frac{x-\alpha}2}\right|$$

Clearly, $\ln\left|-\cot\frac\alpha2\right|$ is independent of $x,$ hence constant

Answer 2

$$\int \dfrac{1}{\sin (x)-\sin(a)}=\int \dfrac{\sin(x)+\sin(a)}{\sin^2 (x)-\sin^2(a)}=\int \dfrac{\sin(x)}{\sin^2 (x)-\sin^2(a)}+\int \dfrac{\sin(a)}{\sin^2 (x)-\sin^2(a)}$$

$$\int \dfrac{\sin(x)}{\sin^2 (x)-\sin^2(a)}=\int \dfrac{\sin(x)}{1-\cos^2 (x)-\sin^2(a)}$$ can be calculated with the substitution $u=\cos(x)$,

$$\int \dfrac{\sin(a)}{\sin^2 (x)-\sin^2(a)}=\sin(a)\int \dfrac{\csc^2(x)}{1-\csc^2(x)\sin^2(a)}=\sin(a)\int \dfrac{\csc^2(x)}{1-\cot^2(x)\sin^2(a)-\sin^2(a)}$$

can be calculated with the substitution $u=\cot(x)$.