What is #f(x) = int xsinx + secxtan^2x -cosx dx# if #f(pi)=-2 #?
1 Answer
Explanation:
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
#J=intsec^3xdx-intsecxdx#
The second integral is known. For
#{(u=secx,=>,du=secxtanx),(dv=sec^2xdx,=>,v=tanx):}#
So:
#J=(secxtanx-intsecxtan^2x)-lnabs(secx+tanx)#
Note that the original integral
#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
#-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
#C=-2-pi#
Then:
#f(x)=1/2secxtanx-1/2lnabs(secx+tanx)-xcosx-pi-2#