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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?

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Why using partial integration? You just can use the formula $[u(x)v(x)]'=u'(x)v(x)+u(x)v'(x)$ – moray95 Jun 1 '13 at 13:45
up vote 1 down vote accepted

$$ f(x) = 2x \cos(x)$$

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.

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$$ 2x\sin \left( x\right) -\int \sin 2dx=2x\sin \left( x\right) -2\left( -\cos \left( x\right) \right) $$ $$ = 2x(sin(x))+2cos(x) $$ $$ = 2(xsin(x))+cos(x)) + c $$

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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. $f(x)=\frac{\mathrm{d}}{\mathrm{d} x}(2x \sin(x))-\sin(x)\frac{\mathrm{d}}{\mathrm{d} x}(2x)=\frac{\mathrm{d}}{\mathrm{d} x}(2x \sin(x))-2\sin(x)$

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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 $

$\Rightarrow I = 2x sinx + cosx +C$

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I learned it as LIATE. comments found here – GEdgar Jun 1 '13 at 14:45

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