8.4.2 - Computing ∫ x² cos(x) dx

[karush]

8.4.2
$$\int {x}^{2}\cos\left({x}\right)\ dx
= {x}^{2}\sin\left({x}\right)
-2\sin\left({x}\right)
+2x\cos\left({x}\right)
+ C
$$
$$uv-\int v \ du $$
$$u={x}^{2}\ \ \ dv=\cos\left({x}\right)\ dx $$
$$du=2x \ dx \ \ \ \ v=\sin\left({x}\right)$$
$${x}^{2}\sin\left({x}\right)-\int\sin\left({x}\right)\ 2x \ dx $$
tried to finish but couldn't get the answer.

 

[Prove It]

Re: 8.4.2

8.4.2
$$\int {x}^{2}\cos\left({x}\right)\ dx
= {x}^{2}\sin\left({x}\right)
-2\sin\left({x}\right)
+2x\cos\left({x}\right)
+ C
$$
$$uv-\int v \ du $$
$$u={x}^{2}\ \ \ dv=\cos\left({x}\right)\ dx $$
$$du=2x \ dx \ \ \ \ v=\sin\left({x}\right)$$
$${x}^{2}\sin\left({x}\right)-\int\sin\left({x}\right)\ 2x \ dx $$
tried to finish but couldn't get the answer.

Use integration by parts again...

 

[karush]

Rewrite $\int\cos\left({x}\right)\ 2x \ dx $ as $2\int x\cos\left({x}\right) \ dx $
Then
$\displaystyle u=x \ \ \ \ \ \ \ dv=\cos\left({x}\right)\ dx $
$\displaystyle du=dx \ \ \ \ \ v=\sin\left({x}\right)$
Then
$$uv-\int v \ du \implies 2\sin\left({x}\right)
-2x\sin\left({x}\right)$$
Finally
$$\int {x}^{2}\cos\left({x}\right)\ dx
= {x}^{2}\sin\left({x}\right)
-2\sin\left({x}\right)
+2x\cos\left({x}\right)
+ C
$$

 

[MarkFL]

We could write:

\(\displaystyle I(x)=\int x^2\cos(x)\,dx\)

Thus, differenting w.r.t $x$, we have the first order linear ODE:

\(\displaystyle \d{I}{x}=x^2\cos(x)\)

We immediately see that the homogeneous solution is:

\(\displaystyle I_h(x)=c_1\)

Then we may assume a particular solution of the form:

\(\displaystyle I_p(x)=(Ax^2+Bx+C)\sin(x)+(Dx^2+Ex+F)\cos(x)\)

Hence:

\(\displaystyle I_p'(x)=(Ax^2+Bx+C)\cos(x)+(2Ax+B)\sin(x)-(Dx^2+Ex+F)\sin(x)+(2Dx+E)\cos(x)=(-Dx^2+(2A-E)x+(B-F))\sin(x)+(Ax^2+(B+2D)x+(C+E))\cos(x)\)

Substituting this into the ODE, we obtain the system:

\(\displaystyle -D=0\)

\(\displaystyle 2A-E=0\)

\(\displaystyle B-F=0\)

\(\displaystyle A=1\)

\(\displaystyle B+2D=0\)

\(\displaystyle C+E=0\)

Solving this system, we obtain:

\(\displaystyle (A,B,C,D,E,F)=(1,0,-2,0,2,0)\)

And so we have:

\(\displaystyle I_p(x)=(x^2-2)\sin(x)+(2x)\cos(x)\)

Hence, by the principle of superposition, we find:

\(\displaystyle I(x)=I_p(x)+I_h(x)=(x^2-2)\sin(x)+2x\cos(x)+C\)

And so we conclude:

\(\displaystyle \int x^2\cos(x)\,dx=(x^2-2)\sin(x)+2x\cos(x)+C\)

 

[karush]

Wow that was interesting
I assume that would be used verses recycling by parts till you get $du=dx$

 

[MarkFL]

Wow that was interesting
I assume that would be used verses recycling by parts till you get $du=dx$

That's a method you will encounter when you study ordinary differential equations...I just posted it to look impressive to be informative. :)

 

[karush]

I swallowed hard ...