We will first start by integrating the expression. Then we'll solve for the constant. Let's start by splitting the integral into two:
#int\ sin(2x)-5sec(x)\ dx=int\ sin(2x)\ dx - int\ 5sec(x)\ dx#
We can evaluate the right integral using this common integral: #int\ sec(x)\ dx=ln|tan(x)+sec(x)|+C#
The left one is a bit more complicated. To get it into an easier form, we're going to use the double angle identity:
#sin(2alpha)=2sin(alpha)cos(alpha)#
Our integral will now look like this:
#int\ 2sin(x)cos(x)\ dx - 5ln|tan(x)+sec(x)|#
Since #d/dx(sin(x))=cos(x)#, we can make a u-substitution so that #u=sin(x)#. To integrate with respect to u, we have to divide by the derivative, which was #cos(x)#:
#int\ (2ucancelcos(x))/cancelcos(x)\ du-5ln|tan(x)+sec(x)|#
#=int\ 2u\ du-5ln|tan(x)+sec(x)|#
#=u^2-5ln|tan(x)+sec(x)|+C#
If we resubstitute, we get:
#=sin^2(x)-5ln|tan(x)+sec(x)|+C#
Now we can use the fact that we know #f(pi/4)=-2# to setup the following equation:
#sin^2(pi/4)-5ln|tan(pi/4)+sec(pi/4)|+C=-2#
If we solve for #C#, we get:
#C=5ln|tan(pi/4)+sec(pi/4)|-sin^2(pi/4)-2#
#tan(pi/4)=1, sec(pi/4)=sqrt2, sin^2(pi/4)=1/2#
#thereforeC=5ln(1+sqrt2)-5/2#
So, the function becomes:
#f(x)=sin^2(x)-5ln|tan(x)+sec(x)|+5ln(1+sqrt2)-5/2#
