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

Question

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

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?

Answer 1

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.

Answer 2

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: $$\int \frac{\cos(u)}{6}du$$

Answer 3

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 :

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