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Solution to a trigonometric problem requested by one student

B.Sudhakar
14/01/2017 0 0

∫ sin³ x / (1 + cos x ) dx multiply Nr & Dr by ( 1 - cos x )

= ∫ sin³ x • ( 1 - cos x ) / ( 1 + cos x ) ( 1 - cos x ) dx 

= ∫ sin³ x • ( 1 - cos x ) / (1 - cos² x ) dx                          ( 1 - cos² x = sin² x )

= ∫ sin³ x • ( 1 - cos x ) / sin² x dx                                    ( sin² x in Dr cancels out )

= ∫ sin x • ( 1 - cos x ) dx 

= ∫ ( sin x - sinx cos x ) dx 

= ∫ sin x d‍x - ∫ sin x cos x dx                        u = cos x,    du = - sin x dx  ∴ dx = - du /sin x

= ∫ sin x dx - ∫ sin x • u ( - 1 / sin x ) du 

= ∫ sin x dx + ∫ u du 

= - cos x + 1/2 u² + C                                   ( substituting u = cos x )

= - cos x + 1/2 cos² x  +C

 

 

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