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

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#