Hi, i'm trying to solve this integral on this site : http://tutorial.math.lamar.edu/Classes/CalcI/SubstitutionRuleIndefinite.aspx. The first question of Example 2:
sin(1-x)(2-cos(1-x))^4 dx and i substituted
u=2-cos(1-x). So i solved for du= sin(1-x)dx but the author of the tutorial says it is du=-sin(1-x)dx.
Since the derivative of cosx is -sinx then the two negatives become positive.
Need some explanation why he still mainstains the negative infront of the sin.
thanks

Question

Hi, i'm trying to solve this integral on this site : http://tutorial.math.lamar.edu/Classes/CalcI/SubstitutionRuleIndefinite.aspx. The first question of Example 2:
 sin(1-x)(2-cos(1-x))^4 dx  and i substituted 
u=2-cos(1-x). So i solved for du= sin(1-x)dx but the author of the tutorial says it is du=-sin(1-x)dx.
Since the derivative of cosx is -sinx then the two negatives become positive.  
Need some explanation why he still mainstains the negative infront of the sin.
thanks

anonymous · Answer

when u take the derivative of 2-cos(1-x)= -(-sin)(1-x)(-1)
derivative of the inside too derivative of (1-x)

anonymous · Answer

that's because the negative come from differentiate of ( 1- x ) that is -1

pokemon23 · Answer

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