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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