What is f(x) = int cos^2x-sec^2x dx if f(pi/8)=-1 ?

1 Answer
Mar 28, 2017

f(x)=1/2x+1/4sin2x-tanx+sqrt2-2-pi/16-1/(4sqrt2)

Explanation:

Note a couple things:

  • cos2x=2cos^2x-1" "=>" "cos^2x=1/2(1+cos2x)
  • d/dxtanx=sec^2x" "=>" "intsec^2xdx=tanx+C

So the integral becomes:

f(x)=intcos^2xdx-intsec^2xdx=1/2int(1+cos2x)dx-tanx

Splitting the remaining integral up:

=1/2intdx+1/2intcos2xdx-tanx

The first integral is simple. For the other, let u=2x so du=2dx. Then:

=1/2x+1/4intcos2x(2dx)-tanx=1/2x+1/2intcosudu-tanx

The integral of cosine is sine:

f(x)=1/2x+1/4sinu-tanx=1/2x+1/4sin2x-tanx+C

We can determine C by using the initial condition f(pi/8)=-1. We could do this with decimals, but for fun we'll do it using identities. Note that tan(x/2)=(1-cosx)/sinx:

-1=1/2(pi/8)+1/4sin(pi/4)-tan(pi/8)+C

-1=pi/16+1/4(1/sqrt2)-(1-cos(pi/4))/sin(pi/4)+C

-1-pi/16-1/(4sqrt2)+(1-1/sqrt2)/(1/sqrt2)=C

C=-1-pi/16-1/(4sqrt2)+(sqrt2-1)

C=sqrt2-2-pi/16-1/(4sqrt2)

Then:

f(x)=1/2x+1/4sin2x-tanx+sqrt2-2-pi/16-1/(4sqrt2)