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

Question

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

Charles Anctil · Accepted Answer

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

Luke Alvarez · Answer

undefined To find $ f(x) = \int x \sin(x) + \sec(x) \tan^2(x) - \cos(x) \, dx $ given $ f(\pi) = -2 $, we'll first integrate each term separately and then apply the given condition $ f(\pi) = -2 $ to solve for the constant of integration.

$$
\begin{align*}
&\int x \sin(x) + \sec(x) \tan^2(x) - \cos(x) \, dx \
&= \left[ -x \cos(x) + \int \cos(x) \, dx \right] + \left[ \int \sec(x) \tan^2(x) \, dx \right] - \sin(x) \
&= -x \cos(x) + \sin(x) + \int \sec(x) \tan^2(x) \, dx
\end{align*}
$$

Now, we need to find $ \int \sec(x) \tan^2(x) \, dx $. We can use substitution method to integrate $ \sec(x) \tan^2(x) $.

Let $ u = \tan(x) $, then $ du = \sec^2(x) \, dx $.

$$
\begin{align*}
\int \sec(x) \tan^2(x) \, dx &= \int u^2 \, du \
&= \frac{u^3}{3} + C \
&= \frac{\tan^3(x)}{3} + C
\end{align*}
$$

Now, we'll substitute this back into our original expression:

$$
\begin{align*}
&-x \cos(x) + \sin(x) + \frac{\tan^3(x)}{3} + C \
\end{align*}
$$

Given that $ f(\pi) = -2 $, we plug in $ x = \pi $ into the expression:

$$
\begin{align*}
-2 &= -\pi \cos(\pi) + \sin(\pi) + \frac{\tan^3(\pi)}{3} + C \
-2 &= \pi - 0 + 0 + C \
C &= -2 - \pi
\end{align*}
$$

So, the expression for $ f(x) $ is:

$$
f(x) = -x \cos(x) + \sin(x) + \frac{\tan^3(x)}{3} - 2 - \pi
$$

Evelyn Agard · Answer

undefined The value of the integral $ \int x \sin(x) + \sec(x) \tan^2(x) - \cos(x) \, dx $ evaluated at $ x = \pi $ is $ -2 $.