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How to integrate 1/sin x

B.Sudhakar
10/01/2017 0 0

∫ 1/sin x dx       ( multiply numerator and denominator by sin x )

= ∫1• sin x / sinx • sinx dx 

= ∫ sin x/ sin² x dx 

= ∫ sin x / (1-cos² x ) dx             (  sin² x + cos² x = 1 )

Let u= cos x.        du = - sin x dx 

= - ∫  sin x /( 1 - u² ) sin x du.    ( sin x cancels out )

= - ∫ 1/( 1 - u² ) du.    

= ∫ 1/( u² - 1 ) du.           ( by changing signs )

= 1/2 [ ∫ 1/(u - 1) - 1/( u + 1 )] du.       [ 1/(u² - 1 ) = 1/2 { 1/(u - 1) - 1/(u + 1) } by partial fraction]

= 1/2 ln | ( u - 1 ) - ( u + 1 ) | +C

= 1/2 ln | (u - 1) / ( u + 1) | + C    ( substitute back u = cos x )

= 1/2 ln | ( cos x - 1) /( cos x + 1) | + C

 

                                     

 

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