What is #f(x) = int xsinx + secxtan^2x -cosx dx# if #f(pi)=-2 #?

Splitting this into three pieces:

#I=intxsinxdx#

Use integration by parts. Let:

#{(u=x,=>,du=dx),(dv=sinxdx,=>,v=-cosx):}#

Then:

#I=-xcosx+intcosxdx=-xcosx+sinx+C#

The next part:

#J=intsecxtan^2x#

Use #tan^2x=sec^2x-1#:

#J=intsec^3xdx-intsecxdx#

The second integral is known. For #sec^3x#, perform integration by parts again, this time letting:

#{(u=secx,=>,du=secxtanx),(dv=sec^2xdx,=>,v=tanx):}#

So:

#J=(secxtanx-intsecxtan^2x)-lnabs(secx+tanx)#

Note that the original integral #J# has reappeared on the right-hand side. Add it to both sides of the equation:

#2J=secxtanx-lnabs(secx+tanx)#

#J=1/2secxtanx-1/2lnabs(secx+tanx)+C#

Finally, we see that the last piece is:

#K=intcosxdx=sinx+C#

Our whole integral is:

#f(x)=I+J-K#

#f(x)=-xcosx+sinx+1/2secxtanx-1/2lnabs(secx+tanx)-sinx+C#

Simplified:

#f(x)=1/2secxtanx-1/2lnabs(secx+tanx)-xcosx+C#

Now solving for #C# using the initial condition #f(pi)=-2#:

#-2=1/2secpitanpi-1/2lnabs(secpi+tanpi)-picospi+C#

#-2=1/2(-1)(0)-1/2lnabs(-1+0)-pi(-1)+C#

#-2=-1/2ln(1)+pi+C#

Since #ln(1)=0#:

#C=-2-pi#

Then:

#f(x)=1/2secxtanx-1/2lnabs(secx+tanx)-xcosx-pi-2#