# Evaluate $\int x\cos (6x) \mathrm dx$?

$$\int x \cos(6x)\, \mathrm dx$$

I have many similar problems to do, but I keep getting stumped on what to do with what resides inside the parenthesis as opposed to an exponent or something in front of the problem say either $\cos^6 (x)$ or $6 \cos (x)$. What should I be doing differently to solve this integral which has the $6x$ evaluated within cosine?

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$\cos(6x)$, $\cos^6(x)$ and $6\cos(x)$ are three different functions. – Michael Albanese Aug 31 '13 at 23:07

## 3 Answers

First do a u-sub on the 6x. The integral then becomes of the form (u)(cosu) with a factor upfront. Then apply Integration By Parts.

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I ended up with 1/6 x sin (6x) + 1/6 cos (6x) + C. Would that be correct? – JAY Aug 31 '13 at 23:24
yes it is correct, worked out above – imranfat Aug 31 '13 at 23:40
@JAY That is not correct; it should be $$\frac{x\sin(6x)}{6}+\frac{\cos(6x)}{36} + C$$ – apnorton Aug 31 '13 at 23:41
@JAY Also, for checking your work, you may like to know about Wolfram Alpha – apnorton Aug 31 '13 at 23:42
I forgot the 3 when typing my solution, this confirms my work. Thank you very much! – JAY Aug 31 '13 at 23:58

integrating by parts : we know $$d(uv) = u \ dv + v \ du$$

$$\Rightarrow \int d(uv) = \int u \ dv + \int v \ du \Rightarrow uv = \int u \ dv + \int v \ du \Rightarrow \int u \ dv = uv - \int v \ du$$

we have the integral $$\int x\cos(6x) \ dx$$

by letting $$u = x \quad , dv = \cos(6x) \ dx$$

$$\Rightarrow du = dx \quad , v = \frac{\sin(6x)}{6}$$

so we integrate by parts and we'll get :

$$\int x\cos(6x) \ dx = uv - \int v \ du = \frac{x\sin(6x)}{6} - \frac{1}{6}\int \sin(6x) \ dx = \frac{x\sin(6x)}{6} + \frac{\cos(6x)}{36} + C$$

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This is what I originally got but my professor had told me that I could not simply leave the 6x, hence my confusion. I used u-substitution before integration by parts and got it in the form 1/36 /u cos (u) du and applied integration by parts afterwards setting u = u and dv = cos (u) du. Does this seem like the right process? – JAY Aug 31 '13 at 23:38
look at this $$y= 6x \Rightarrow x = \frac{y}{6} \Rightarrow \frac{1}{6}dy = dx$$ $$\Rightarrow I = \frac{1}{36}\int y\cos( y) \ dy$$ $$= \frac{1}{36} \left( y\sin(y) - \int \sin y \ dy \right)$$ $$= \frac{1}{36} \left(y\sin(y) + \cos y + c\right)$$ $$= \frac{1}{36} \left( 6x\sin(6x) + \cos(6x) + c \right)$$ $$= \frac{x\sin(6x)}{6} + \frac{\cos(6x)}{36} + c_1$$ doesn't change any thing – what'sup Aug 31 '13 at 23:44

Whenever you have the product of:

1. Something you know how to differentiate (e.g. $x$), and...
2. Something you know how to integrate (e.g. $\cos(6x)$)

...you should use integration by parts.

For evaluating $\int \cos(6x)\,dx$, we use a $u$-substitution; let $u=6x$. This means $du = 6\,dx$. Now we have: $$\int \frac{\cos(u)}{6}du$$

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