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