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

1 Answer
Mar 26, 2016

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