Question

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

Answer 1

`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`

Answer 2

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

Explanation:

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