Question

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

Answer 1

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

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 `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`