Can anyone please explain this questionto me. This is urgent

Dear student

 

I = ∫ 1/ √[ sin³ x. sin (x+a) ] dx 

= ∫ 1 / { sin x. √[ sin x.( sin x. cos a + cos x. sin a )] } dx 

= ∫ csc x / √[ sin² x.(cos a + (cos x / sin x). sin a) ] dx 

= ∫ csc x / [ sin x.√( cos a + cot x. sin a ) ] dx 

= ∫ csc² x / √( cos a + cot x. sin a ) dx ..................... (1) 
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Let : u = cos a + cot x. sin a 

Then : du/dx = - sin a. csc² x 

so that : csc² dx = - csc a du. 
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From (1), then, 

I = ∫ ( 1 / √u ) ( - csc a ) du 

= - csc a. ( 2√u ) + C 

= - 2. csc a. √( cos a + cot x. sin a ) + C ... Ans.

 

hope it helps

 

Regards