What is f(x) = int x-sin2x+cosx dx if f(pi/2)=3 ?
2 Answers
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
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
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
Explanation:
Le us find
Given
Substituting
Therefore,