What is #F(x) = int sin2xcos^2x-sinxcos^2x dx# if #F(pi) = 1 #?

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Answer 1 · 2017-01-09T14:05:55

#F(x) = (cos^3x)/3- (cos^4x)/2+11/6#

Let's calculate the indefinite integral:

#int (sin2xcos^2x-sinxcos^2x)dx = int cos^2x(sin2x-sinx)dx #

Using the identity:

#sin2x = 2 sinxcosx#

#int cos^2x(sin2x-sinx)dx = int cos^2x(2sinxcosx-sinx)dx=int cos^2xsinx(2cosx-1)dx#

As #sinxdx = -d(cosx)#

#int cos^2xsinx(2cosx-1)dx = int (cos^2x-2cos^3x)d(cosx) = (cos^3x)/3- (cos^4x)/2+C#

Now we determine the constant from:

#F(pi) = 1#

#(cos^3pi)/3- (cos^4pi)/2+C = 1#

#(-1)^3/3-(-1)^4/2 +C = 1#

#-1/3-1/2 +C =1#

#C = 1+5/6=11/6#

So, finally:

#F(x) = (cos^3pi)/3- (cos^4pi)/2+11/6#