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

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

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

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

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

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


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

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 $$ \frac{\cos(x/4)}{x}$$

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My friend Kiyoshi is blind. He can't read your scan. My friend John is dyslexic. He can't read it either. These are among the reasons we urge users to format their posts using MathJax. You can start learning how to use it by reading this post. – dfeuer Dec 9 '13 at 9:31
Il have a look at that now, in the meantime, what Im asking is how to integrate cos(x/4)/x – Samir Chahine Dec 9 '13 at 9:34
I edited my question a little bit, Have a look and see if it kinda of makes sense, sorry for the hassle. – Samir Chahine Dec 9 '13 at 9:39
up vote 2 down vote accepted

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*}

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my final answer was that, except I left out the $x$ and i had $\frac{48}{x} sin..$ Stupid mistakes, I'l have another go at the question and Hopefully get it, thank you! – Samir Chahine Dec 9 '13 at 9:57

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