Question

What is `f(x) = int 3x-3sec(x/2) dx` if `f((7pi)/4) = 0`?

Answer 1

`f(x)=(3x^2)/2-6lnabs(sec(x/2)+tan(x/2))-42.9193`

Explanation:

Split the integral into two simpler integral expressions:

`f(x)=3intxdx-3intsec(x/2)dx`

The first integral can be integrated using the rule:

`intx^ndx=x^(n+1)/(n+1)+C`

Which gives

`=3(x^2/2)-3intsec(x/2)dx`

We don't yet need to add `C`, the constant of integration, until we find the final integral.

As for the remaining integral, we will want to get it into the form:

`intsec(u)du=lnabs(sec(u)+tan(u))+C`

Thus, we will set `u=x/2`, so `du=1/2dx`.

We will multiply the interior of the integral by `1/2` and the exterior by `2` to balance the equality.

`=(3x^2)/2-3(2)intsec(x/2)*1/2dx`

Substitute in `u` and `du`:

`=(3x^2)/2-6intsec(u)du+C`

This matches our integral identity:

`=(3x^2)/2-6lnabs(sec(u)+tan(u))+C`

Substitute `x/2` for `u`:

`f(x)=(3x^2)/2-6lnabs(sec(x/2)+tan(x/2))+C`

We can now use the original condition that `f((7pi)/4)=0` to solve for `C`.

`0=(3((7pi)/4)^2)/2-6lnabs(sec((7pi)/8)+tan((7pi)/8))+C`

It would be a horrendous mess to try to evaluate this without decimals, since we would need to use half-angle formulas to "simplify," so use a calculator to find these values:

`0=45.3385-2.4192+C`

`C=-42.9193`

This gives us the function:

`f(x)=(3x^2)/2-6lnabs(sec(x/2)+tan(x/2))-42.9193`