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

2 Answers
Feb 4, 2017

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

Explanation:

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

Feb 4, 2017

f(x)=x^2/2+1/2cos2x+sinx+(20-pi^2)/8

Explanation:

f(x)=intx-sin2x+cosxdx
" "
f(x)=intxdx-intsin2xdx+intcosxdx+C " "Cis a constant
" "
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