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∫ sin² x /( 1+ cos x ) dx
Let's multiply both numerator and denominator by ( 1 - cos x )
= ∫ sin² x • ( 1 - cos x ) / ( 1 + cos x ) • ( 1 - cos x ) dx
= ∫ sin² x • (1 - cos x ) / ( 1 - cos² x ) dx
= ∫ sin² x • ( 1 - cos x ) / ( sin² x ) dx ( 1 - cos² x = sin² x )
= ∫ ( 1 - cos x ) dx ( sin² x cancels out )
= ∫ 1• dx - ∫ cos x dx
= x - sin x + C
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