Question #3ee60

1 Answer
Gaurav
Aug 10, 2014

#=-(cos3x)/6-(cosx)/2+c#, where #c# is a constant

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Explanation

#=intsin2xcosxdx#

From trigonometric identities,

#sinAcosB=1/2(sin(A+B)+sin(A-B))#

Similarly,
#sin2xcosx=1/2(sin3x+sinx)#

#=intsin2xcosxdx=int1/2(sin3x+sinx)dx#

#=1/2intsin3xdx+1/2intsinxdx#

#=1/2((-cos3x)/3)+1/2(-cosx)+c#, where #c# is a constant

#=-(cos3x)/6-(cosx)/2+c#, where #c# is a constant

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