calculus - Evaluating: $int 3xsinleft(frac x4right) , dx$.

$\displaystyle\color{darkblue}{3\int x\sin\left(\dfrac x4\right)\,\mathrm dx}$

$$\begin{align} \dfrac{-4x}{x}\cos\dfrac x4 \,\,\boldsymbol\Rightarrow\,\, & -4\cos\left(\dfrac x4\right)-\int \dfrac{-4}{x}\cos\left(\dfrac x4 \right)\,\mathrm dx\\\,\\ & 3\left(-4\cos\left(\dfrac x4\right)+\int\dfrac4x\cos\left(\dfrac x4\right)\,\mathrm dx\right)\\\,\\ &\int\dfrac{4\cos\left(\frac x4\right)}{x}\mathrm dx \end{align}$$ $\displaystyle \color{darkblue}{uv-\int v\dfrac{\mathrm du}{\mathrm dx}\,\mathrm dx }$

$\displaystyle\boxed{\displaystyle\,\,-4\cos\left(\dfrac x4\right)+4\int\dfrac{\cos x/4}{x}\,\mathrm dx\,\,}$

$\displaystyle 3\left[-4\cos\left( \dfrac x4\right) +4\left(\cos\left( \dfrac x4\right)\ln(x)\right)\right]$

$\displaystyle -12\cos (x/4) + 12\cos (x/4) \ln(x) \rightarrow \text{ wrong.}$

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$\displaystyle\color{darkblue}{\int 3x\sin\left(\dfrac x4\right)\,\mathrm dx}$
$\quad\quad\quad\quad\displaystyle\int\dfrac{\cos(x/4)}{x}\rightarrow\dfrac1x\int\cos(x/4)\mathrm dx$

$\displaystyle3\left[-4\cos(x/4)+\dfrac4x\sin(x/4)\right]$

$\displaystyle -12\cos(x/4)+\dfrac{48}{x}\sin(x/4)$

In the text above is my work done to solve the following question:

Find the indefinite integral of: $\left[3x\sin\left(\dfrac x4\right)\right]$

The bordered area is the furthest I got (there should be a 3 at the front to multiply the whole equation but I usually remember to add that at the end) The part where I wrote "wrong" is what I thought the answer was, I assumed it would be such. What I'm having troubles with is integrating the $\frac{\cos(x/4)}{x}$. Would I need to make it $\frac{\cos(x/4)}{x}$ and then integrate by parts to get that integral? Thanks in advance, I hope I made some sense in what I'm trying to achieve. I guess what I'm looking for, is a way to integrate $$\frac{\cos(x/4)}{x}$$

Answer

I think you have made a mistake in the application of integration by parts. Taking $u$ as $x$ and $\frac{\mathrm{d}v}{\mathrm{d}x}$ as $\sin \frac{x}{4}$, we should get
$$\begin{align*} 3 \int {x} \sin \frac{x}{4} \mathrm{d}x &= 3 ( -4x \cos \frac{x}{4}- \int -4 \cos \frac{x}{4} \mathrm{d}x) \\ &= -12x \cos \frac{x}{4} + 48 \sin \frac{x}{4} + C. \end{align*}$$