What is #f(x) = int x-sin2x+cosx dx# if #f(pi/2)=3 #?

We will use the following rules:

• #intf(x)+-g(x)dx=intf(x)dx+-intg(x)dx#
• #intx^ndx=x^(n+1)/(n+1)+C#
• #intsin(u)du=-cos(u)+C#
• #intcos(u)du=sin(u)+C#

So:

#f(x)=intx^1dx-intsin(2x)dx+intcos(x)dx#

The first and third can be found directly:

#f(x)=x^2/2-intsin(2x)dx+sin(x)#

For the remaining integral, let #u=2x# so #du=(2)dx#. Then:

#f(x)=x^2/2-1/2intsin(2x)(2)dx+sin(x)#

#f(x)=x^2/2-1/2intsin(u)du+sin(x)#

#f(x)=x^2/2-1/2(-cos(u))+sin(x)+C#

#f(x)=x^2/2+1/2cos(2x)+sin(x)+C#

We can now solve for #C# using #f(pi/2)=3#:

#3=(pi/2)^2/2+1/2cos(2*pi/2)+sin(pi/2)+C#

#3=pi^2/8+1/2(-1)+1+C#

#C=5/2-pi^2/8#

So:

#f(x)=x^2/2+1/2cos(2x)+sin(x)+5/2-pi^2/8#

#f(x)=intx-sin2x+cosxdx#
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#f(x)=intxdx-intsin2xdx+intcosxdx+C# #" "C#is a constant
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#f(x)=x^2/2-1/2intd(-cos2x)+intd(sinx)+C#
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#f(x)=x^2/2+1/2cos2x+sinx+C#
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Le us find #C#
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Given #f(pi/2)=3#
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#f(pi/2)=(pi/2)^2/2+1/2cos(2*pi/2)+sin(pi/2)+C#
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Substituting #f(pi/2)=3#
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#rArr3=(pi)^2/8+1/2cospi+1+C#
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#rArr3=(pi)^2/8-1/2+1+C#
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#rArr3=pi^2/8+1/2+C#
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#rArrC=3-1/2-pi^2/8#
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#rArrC=24/8-4/8-pi^2/8#
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#rArrC=(20-pi^2)/8#
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Therefore,#f(x)=x^2/2+1/2cos2x+sinx+(20-pi^2)/8#