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)#
