MHB How is Integration by Parts Applied to $\int_{0}^{\pi} x^3 \cos(x) \, dx$?

[karush]

$\Large {S6-7.1.24}$
$$
\displaystyle
I=\int_{0}^{\pi} {x}^{3}\cos\left({x}\right)\,dx=12-3{\pi}^{2} \\
\begin{align}
u& = {{x}^{3}} & dv&=\cos\left({x}\right) \, dx \\
du&={3x^2} \ d{x}& v&={\sin\left({x}\right)}
\end{align} \\
$$
$$
\text{IBP} \displaystyle =uv-\int v\ du \\
\displaystyle{x}^{3}\cdot \sin\left({x}\right)
-\int \sin\left({x}\right) \cdot 3{x}^{2} \ d{x} \\
\begin{align}
u& = {{3x}^{2}} & dv&=\sin\left({x}\right) \, dx \\
du&={6x} \ d{x}& v&={-\cos\left({x}\right)}
\end{align} \\
$$
$$
\displaystyle -3{x}^{2}\cdot \cos\left({x}\right)
+6\int \cos\left({x}\right) \cdot {x} \ d{x} \\
\begin{align}
u& = {x} & dv&=\cos\left({x}\right) \, dx \\
du&={dx} \ d{x}& v&={\sin\left({x}\right)}
\end{align}
$$
$
\text{continued with this one more IBP but didn't get the answer}
$

 

[MarkFL]

This is how I would work the problem...

We are given:

$$I=\int_0^{\pi} x^3\cos(x)\,dx$$

Using IBP, I agree the choices should be:

$$u=x^3\,\therefore\,du=3x^2\,dx$$

$$dv=\cos(x)\,dx\,\therefore\,v=\sin(x)$$

And we have:

$$I=\left[x^2\sin(x)\right]_0^\pi-3\int_0^{\pi} x^2\sin(x)\,dx=-3\int_0^{\pi} x^2\sin(x)\,dx$$

Using IBP again, where:

$$u=x^2\,\therefore\,du=2x\,dx$$

$$dv=\sin(x)\,dx\,\therefore\,v=-\cos(x)$$

And we have:

$$I=3\left(\left[x^2\cos(x)\right]_0^{\pi}-2\int_0^{\pi} x\cos(x)\,dx\right)=-3\left(\pi^2+2\int_0^{\pi} x\cos(x)\,dx\right)$$

Using IBP again, where:

$$u=x\,\therefore\,du=dx$$

$$dv=\cos(x)\,dx\,\therefore\,v=\sin(x)$$

And we have:

$$I=-3\left(\pi^2+2\left(\left[x\sin(x)\right]_0^{\pi}-\int_0^{\pi} \sin(x)\,dx\right)\right)=-3\left(\pi^2-2\int_0^{\pi} \sin(x)\,dx\right)=-3\left(\pi^2+2\left[\cos(x)\right]_0^{\pi}\right)=3\left(4-\pi^2\right)=12-3\pi^2$$

 

[karush]

thank you, I see where i went astray didn't factor correctly.
i seem to be ok with what to do i get killed with simple arithmetic.

so when we see a $x^n$ its a hint how many IBP we have to do