Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How to integrate :

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

share|improve this question
    
Welcome to Math.SE. What methods have you tried? –  Nicholas R. Peterson Jun 14 '13 at 15:06
    
Hello, welcome to Math.SE. I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance. –  Lord_Farin Jun 14 '13 at 15:22

2 Answers 2

up vote 5 down vote accepted

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

share|improve this answer

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

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.