Partial integration We want to integrate 
$$ f(x) = 2x \cos(x)$$
We use partial integration where $2x = g$ and $\cos(x) = f'$
I end up with 
$$d (2x \sin(x)) - \sin(x) \cdot d2x$$
What confuses me is the term $\sin(x) \cdot d2x$
How would I be able to evaluate this term?
 A: 
$$ f(x) = 2x \cos(x)$$
We use partial integration where $2x = g$ and $\cos(x) = f'$

I know what you mean but if $ f(x) = 2x \cos(x)$, $f'(x)$ is not equal to $\cos x$. So I denote $$u(x)=2x,\qquad v^{\prime }(x)=\cos x.$$
Then 
$$
\begin{eqnarray*}
u^{\prime }(x) &=&2 \\
v(x) &=&\int \cos x\ dx=\sin x.
\end{eqnarray*}
$$
From
$$
\begin{equation*}
I=\int u(x)v^{\prime }(x)\ dx=u(x)v(x)-\int u^{\prime }(x)v(x)\ dx,
\end{equation*}
$$
we get  
$$
\begin{eqnarray*}
I &=&\int \underset{u(x)}{\underbrace{2x}}\cdot\text{ }\underset{v^{\prime }(x)}{\underbrace{\cos x}}dx=\underset{u(x)}{\underbrace{2x}}\cdot\underset{u(x)}{\underbrace{\sin x}}-\int \underset{u'(x)}{\underbrace{2}}\cdot\underset{v(x)}{\underbrace{\sin x}}\ dx \\
&=&2x\sin x-(-2\cos x)+C \\
&=&2x\sin x+2\cos x+C.
\end{eqnarray*}
$$

What confuses me is the term $\sin(x) \cdot d2x$

The term we got was $\sin(x) \cdot 2x$.
Added: We have selected $u(x)$ and $v(x)$ according to the LIATE rule (L ogarithm I nverse trigonometric A lgebraic T rigonometric E xponential). By doing this, we try to:


*

*Find easily $v(x)$ from $v'(x)$, and 

*Evaluate $\int v(x)u'(x)\ dx$ easier than $\int u(x)v'(x)\ dx$.


This rule works most of the times because when we differentiate the polynomial (algebraic function) $u(x)=2x$ we get a simpler algebraic function and when we integrate the direct trigonometric function $v'(x)=\cos x$ we get another direct trigonometric function.
A: $$ 2x\sin \left( x\right) -\int \sin 2dx=2x\sin \left( x\right) -2\left( -\cos \left( x\right) \right) $$
$$ = 2x(sin(x))+2cos(x) $$
$$ = 2(xsin(x))+cos(x)) + c $$
A: You may find this makes more sense to you if you include a $\mathrm{d} x$ term in the denominator of each term i.e. $f(x)=\frac{\mathrm{d}}{\mathrm{d} x}(2x \sin(x))-\sin(x)\frac{\mathrm{d}}{\mathrm{d} x}(2x)=\frac{\mathrm{d}}{\mathrm{d} x}(2x \sin(x))-2\sin(x)$
A: While doing integration by parts use : ILATE ( I - inverse trigo function; L - logarithmic function; A - algebric function ; T - trigonometric function , E - exponential function ) to choose first function and second function so, in your case : 1st function can be taken as $ f(x) = 2x $ and second function can be taken as $g(x) = cosx$
Therefore using formula of integration by parts which is $ I = f(x) \int g(x) -\int\{\frac{d g(x)}{dx} \int f(x)\}$
$\Rightarrow I = 2x \int cos(x) -\int \{\frac{d (x)}{dx}\int cos(x) dx\}$
$\Rightarrow I = 2x sinx -\int \{1. sin(x) dx\}$
$\Rightarrow I = 2x sinx - (-cosx) +C $
$\Rightarrow I = 2x sinx + cosx +C$ 
