What is f(x) = int sin2x-5secx dx if f(pi/4)=-2 ?

1 Answer
Nov 27, 2017

f(x)=sin^2(x)-5ln|tan(x)+sec(x)|+5ln(1+sqrt2)-5/2

Explanation:

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