Question

Integrate √(sin(x-a)/sin(x-b)) wrt x.

Answer 1

This problem is tough and the answer comes in a complex form.

Integrate  √ [(sin (x-a) / Sin(x-b) ]  dx

Let  c = b-a,   x-b = y,    x-a = y+c
Let  t² = sin(x-a) / Sin(x-b) = sin (y+c) / sin y          --- (1)
       2t dt  = - sin c /sin² y   dy                                --- (2)

Expand (1) and square both sides to get:
     sin²y = sin²c /(t⁴ -2cos c t²+1) = sin² c/(t⁴ -2t²+1 

f(x) dx
 = -2 sin c * t² dt/[t⁴ + 2t² +1 - 4 t² cos² c/2]   
 = - 2 sin c * t² dt / [(t²+1+ 2t cos c/2) (t² + 1 - 2t cos c/2) ]  
 = -2 sin c * 1/(4 cos c/2)*[ t dt/ (t² - 2t cos c/t +1)  -  t dt/(t² +2t cos c/2 +1) ]
 = - sin c/2 * [ t dt /[ (t - cos c/2)² + sin² c/2]  -  t dt/ [ (t+cos c/2)² + sin² c/2 ]  ]
 = - sin c/2 *  z dz / (1+ z²)  - cos c/2 dz /(1+z²)
             + sin c/2 * s ds /(1+s²)  -  cos c/2 * ds /(1+s²)

where  z = (t - cos c/2) / sin c/2   , dt = dz sin c/2
            s = (t + cos c/2) / sin c/2  , dt = ds sin c/2

When we integrate we get:
ans = - 1/2 *sin c/2 * Ln [(t² +2 t cos c/2 +1) / (t² - 2t cos c/2 +1) ]
               - cos c/2 * Tan⁻¹  [ 2 t sin c/2 / (1 -t²) ]  + K

Substitute for t , c, and t² to get the integral as a function of x.

Answer 2

Answer:

sinb-a ln sinx-b+x-bcosb-a+C  

Step-by-step explanation:

Let I=∫sin(x−a)sin(x−b)dx

Putting x−b=t  

⇒  x=b+t & dx=dt

∴ I= ∫sin(b+t−a)sintdt    

=∫sin{(b−a)+t}sintdt    

= ∫sin(b−a)cos tsint+∫cos(b−a) sin tsin tdt      

= ∫sin(b−a)cot t dt+∫cos(b−a)dt      

= sin(b−a) ln |sin t|+t cos(b−a)+C      

= sin(b−a) ln |sin(x−b)|+(x−b)cos(b−a)+C        [∵t=x−b]  

Let I=∫sinx-asinx-bdx

Putting x-b=t  

⇒  x=b+t& dx=dt

∴ I= ∫sinb+t-asintdt      

=∫sinb-a+tsintdt      

= ∫sinb-acos tsint+∫cosb-a sin tsin tdt    

= ∫sinb-acot t dt+∫cosb-adt    

= sinb-a ln sin t+t cosb-a+C      

= sinb-a ln sinx-b+x-bcosb-a+C      

∵t=x-b