Integrating $-3x\cos{4x}$: Step-by-Step

[karush]

$\tiny{9.1}$
\begin{align*} \displaystyle
I&=\int -3x \cos{4x} \, dx = -3\int x \cos{4x} \, dx\\
\textit{uv substitution}\\
u&=x\therefore du=dx\\
dv&=\cos{4x} \, dx\therefore v=-\frac{\sin{4x}}{4}\\
\textit{now we have}\\
I&=3\left[\frac{x\sin{4x}}{4}
+\int\frac{\sin{4x}}{4} \, dx\right]\\
\textit{and finally}\\
I&=\frac{3}{4} x\sin{4x} +\frac{1}{16}\cos{4x}+C
\end{align*}

think this is ok but probably suggestions

 

[MarkFL]

Let's check your result using a different method (one you'll soon learn)...let:

\(\displaystyle \d{I}{x}=-3x\cos(4x)\)

We know the homogeneous solution will be a constant, $I_h=c_1$.

Using the method of undetermined coefficients, we will expect a particular solution of the form:

\(\displaystyle I_p=(Ax+B)\cos(4x)+(Cx+D)\sin(4x)\)

Differentiating w.r.t $x$, we obtain:

\(\displaystyle \d{I}{x}=(Ax+B)(-4\sin(4x))+A\cos(4x)+(Cx+D)(4\cos(4x))+C\sin(4x)=(4Cx+4D+A)\cos(4x)+(-4ax-4B+C)\sin(x)\)

Substituting that into our ODE, we find:

\(\displaystyle (4Cx+4D+A)\cos(4x)+(-4Ax-4B+C)\sin(x)=(-3x+0)\cos(4x)+(0x+0)\sin(4x)\)

Equating coefficients, there results:

\(\displaystyle 4C=-3\)

\(\displaystyle 4D+A=0\)

\(\displaystyle -4A=0\)

\(\displaystyle -4B+C=0\)

From this, we obtain:

\(\displaystyle (A,B,C,D)=\left(0,-\frac{3}{16},-\frac{3}{4},0\right)\)

Hence:

\(\displaystyle I_p=\left(0x-\frac{3}{16}\right)\cos(4x)+\left(-\frac{3}{4}x+0\right)\sin(4x)=-\frac{3}{16}\left(4x\sin(4x)+\cos(4x)\right)\)

So, by the principle of superposition, the general solution is:

\(\displaystyle I=I_h+I_p=-\frac{3}{16}\left(4x\sin(4x)+\cos(4x)\right)+c_1\)

It looks like you made only a couple of very minor slips. Can you spot them?

 

[karush]

left $dx$ off

always get lost with signs

that was an interesting method

im going to do another IBP

I took a test thursday but ran out of time half way!
to much time figuring next step

 

[Prove It]

$\tiny{9.1}$
\begin{align*} \displaystyle
I&=\int -3x \cos{4x} \, dx = -3\int x \cos{4x} \, dx\\
\textit{uv substitution}\\
u&=x\therefore du=dx\\
dv&=\cos{4x} \, dx\therefore v=-\frac{\sin{4x}}{4}\\
\textit{now we have}\\
I&=3\left[\frac{x\sin{4x}}{4}
+\int\frac{\sin{4x}}{4} \, dx\right]\\
\textit{and finally}\\
I&=\frac{3}{4} x\sin{4x} +\frac{1}{16}\cos{4x}+C
\end{align*}

think this is ok but probably suggestions

Another option - start by differentiating your function twice

$\displaystyle \begin{align*} \frac{\mathrm{d}^2}{\mathrm{d}x^2}\,\left[ -3\,x\cos{ \left( 4\,x \right) } \right] &= \frac{\mathrm{d}}{\mathrm{d}x} \,\left[ 12\,x\sin{ \left( 4\,x \right) } - 3 \cos{ \left( 4\,x \right) } \right] \\ &= 48\,x\cos{ \left( 4\,x \right) } + 12\sin{ \left( 4\,x \right) } + 12 \cos{ \left( 4\,x \right) } \end{align*}$

and since every derivative equation can be written in an equivalent form as an integral, as $\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} \,\left[ 12\,x\sin{ \left( 4\,x \right) } - 3\cos{ \left( 4\,x \right) } \right] &= 48\,x\cos{ \left( 4\,x \right) } + 12\sin{ \left( 4\,x \right) } + 12\cos{ \left( 4\,x \right) } \end{align*}$ that means

$\displaystyle \begin{align*} \int{ \left[ 48\,x\cos{ \left( 4\,x \right) } + 12\sin{ \left( 4\,x \right) } + 12\cos{ \left( 4\,x \right) } \right] \,\mathrm{d}x } &= 12\,x\sin{ \left( 4\,x \right) } - 3\cos{ \left( 4\,x \right) } + C_1 \\ \int{ 48\cos{ \left( 4\,x \right) } \,\mathrm{d}x} + \int{12\sin{ \left( 4\,x \right) } \,\mathrm{d}x} + \int{ 12\cos{ \left( 4\,x \right) } \,\mathrm{d}x} &= 12\,x\sin{ \left( 4\,x \right) } - 3\cos{ \left( 4\,x \right) } + C_1 \\ \int{48\,x\cos{ \left( 4\,x \right) } \,\mathrm{d}x} - 3\cos{ \left( 4\,x \right) } + C_2 + 3\sin{ \left( 4\,x \right) } + C_3 &= 12\,x \sin{ \left( 4\,x \right) } - 3\cos{ \left( 4\,x \right) } + C_1 \\ \int{ 48\,x\cos{ \left( 4\,x \right) }\,\mathrm{d}x} &= 12\,x\sin{ \left( 4\,x \right) } - 3\cos{ \left( 4\,x \right) } + 3\cos{ \left( 4\,x \right) } - 3\sin{ \left( 4\,x \right) } + C_1 - C_2 - C_3 \\ -16 \int{ -3\,x\cos{ \left( 4\,x \right) } \,\mathrm{d}x } &= 12\,x\sin{ \left( 4\,x \right) } - 3\sin{ \left( 4\,x \right) } + C_1 - C_2 - C_3 \\ \int{ -3\,x\cos{ \left( 4\,x \right) } \,\mathrm{d}x} &= \frac{3}{16}\,\sin{ \left( 4\,x \right) } - \frac{3}{4}\,x\sin{ \left( 4\,x \right) } + C \textrm{ where } C = -\frac{1}{16}\,\left( C_1 - C_2 - C_3 \right) \end{align*}$

 

[karush]

lots of steps!

 

[I like Serena]

$\tiny{9.1}$
\begin{align*} \displaystyle
I&=\int -3x \cos{4x} \, dx = -3\int x \cos{4x} \, dx\\
\textit{uv substitution}\\
u&=x\therefore du=dx\\
dv&=\cos{4x} \, dx\therefore v=-\frac{\sin{4x}}{4}\\
\textit{now we have}\\
I&=3\left[\frac{x\sin{4x}}{4}
+\int\frac{\sin{4x}}{4} \, dx\right]\\
\textit{and finally}\\
I&=\frac{3}{4} x\sin{4x} +\frac{1}{16}\cos{4x}+C
\end{align*}

think this is ok but probably suggestions

lots of steps!

Less steps... and no mistakes with minus signs with a couple of sub steps... (Thinking)
$$I = \int -3x \cos{4x} \, dx
= -\frac 34 \int x\, d(\sin{4x})
\overset{\boxed{\int udv =uv -\int vdu}}= -\frac 34 \left(x\sin 4x - \int \sin 4x\,dx\right) \\
= -\frac 34 \left(x\sin 4x + \frac 14 \cos 4x\right) + C
= -\frac 34 x\sin 4x - \frac 3{16} \cos 4x + C
$$