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Formula
Calculate the integral
$$\displaystyle\int { x \cos\left( x \right) } d { x }$$
$1 \left ( x \sin\left( x \right) - \left ( - \cos\left( x \right) \right ) \right )$
Calculate the integral
$\displaystyle\int { \color{#FF6800}{ x } \color{#FF6800}{ \cos\left( \color{#FF6800}{ x } \right) } } d { \color{#FF6800}{ x } }$
 Calculate the integral using the formula of $\int x \cos^{n}(x) dx =\dfrac{1}{n}(x\sin(x)\cos^{n-1}(x)-(\int{\sin(x)\cos^{n-1}(x)}d{x})+(n-1)\int{x\cos^{n-2}(x)}d{x})$
$\color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 1 } } } \left ( \color{#FF6800}{ x } \color{#FF6800}{ \sin\left( \color{#FF6800}{ x } \right) } \color{#FF6800}{ \cos ^ { \color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } } \left ( \color{#FF6800}{ x } \right) } \color{#FF6800}{ - } \left ( \displaystyle\int { \color{#FF6800}{ \sin\left( \color{#FF6800}{ x } \right) } \color{#FF6800}{ \cos ^ { \color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } } \left ( \color{#FF6800}{ x } \right) } } d { \color{#FF6800}{ x } } \right ) \color{#FF6800}{ + } \left ( \color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) \displaystyle\int { \color{#FF6800}{ x } \color{#FF6800}{ \cos ^ { \color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ 2 } } \left ( \color{#FF6800}{ x } \right) } } d { \color{#FF6800}{ x } } \right )$
$\dfrac { 1 } { 1 } \left ( x \sin\left( x \right) \cos ^ { \color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } } \left ( x \right) - \left ( \displaystyle\int { \sin\left( x \right) \cos ^ { 1 - 1 } \left ( x \right) } d { x } \right ) + \left ( 1 - 1 \right ) \displaystyle\int { x \cos ^ { 1 - 2 } \left ( x \right) } d { x } \right )$
 Add $1$ and $- 1$
$\dfrac { 1 } { 1 } \left ( x \sin\left( x \right) \cos ^ { \color{#FF6800}{ 0 } } \left ( x \right) - \left ( \displaystyle\int { \sin\left( x \right) \cos ^ { 1 - 1 } \left ( x \right) } d { x } \right ) + \left ( 1 - 1 \right ) \displaystyle\int { x \cos ^ { 1 - 2 } \left ( x \right) } d { x } \right )$
$\dfrac { 1 } { 1 } \left ( x \sin\left( x \right) \color{#FF6800}{ \cos ^ { \color{#FF6800}{ 0 } } \left ( \color{#FF6800}{ x } \right) } - \left ( \displaystyle\int { \sin\left( x \right) \cos ^ { 1 - 1 } \left ( x \right) } d { x } \right ) + \left ( 1 - 1 \right ) \displaystyle\int { x \cos ^ { 1 - 2 } \left ( x \right) } d { x } \right )$
 Calculate power 
$\dfrac { 1 } { 1 } \left ( x \sin\left( x \right) \times \color{#FF6800}{ 1 } - \left ( \displaystyle\int { \sin\left( x \right) \cos ^ { 1 - 1 } \left ( x \right) } d { x } \right ) + \left ( 1 - 1 \right ) \displaystyle\int { x \cos ^ { 1 - 2 } \left ( x \right) } d { x } \right )$
$\dfrac { 1 } { 1 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ \sin\left( \color{#FF6800}{ x } \right) } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } - \left ( \displaystyle\int { \sin\left( x \right) \cos ^ { 1 - 1 } \left ( x \right) } d { x } \right ) + \left ( 1 - 1 \right ) \displaystyle\int { x \cos ^ { 1 - 2 } \left ( x \right) } d { x } \right )$
 Multiplying any number by 1 does not change the value 
$\dfrac { 1 } { 1 } \left ( \color{#FF6800}{ x } \color{#FF6800}{ \sin\left( \color{#FF6800}{ x } \right) } - \left ( \displaystyle\int { \sin\left( x \right) \cos ^ { 1 - 1 } \left ( x \right) } d { x } \right ) + \left ( 1 - 1 \right ) \displaystyle\int { x \cos ^ { 1 - 2 } \left ( x \right) } d { x } \right )$
$\dfrac { 1 } { 1 } \left ( x \sin\left( x \right) - \left ( \displaystyle\int { \sin\left( x \right) \cos ^ { \color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } } \left ( x \right) } d { x } \right ) + \left ( 1 - 1 \right ) \displaystyle\int { x \cos ^ { 1 - 2 } \left ( x \right) } d { x } \right )$
 Add $1$ and $- 1$
$\dfrac { 1 } { 1 } \left ( x \sin\left( x \right) - \left ( \displaystyle\int { \sin\left( x \right) \cos ^ { \color{#FF6800}{ 0 } } \left ( x \right) } d { x } \right ) + \left ( 1 - 1 \right ) \displaystyle\int { x \cos ^ { 1 - 2 } \left ( x \right) } d { x } \right )$
$\dfrac { 1 } { 1 } \left ( x \sin\left( x \right) - \left ( \displaystyle\int { \color{#FF6800}{ \sin\left( \color{#FF6800}{ x } \right) } \color{#FF6800}{ \cos ^ { \color{#FF6800}{ 0 } } \left ( \color{#FF6800}{ x } \right) } } d { \color{#FF6800}{ x } } \right ) + \left ( 1 - 1 \right ) \displaystyle\int { x \cos ^ { 1 - 2 } \left ( x \right) } d { x } \right )$
 Substitute with $u = \cos\left( x \right)$ and calculate the integral 
$\dfrac { 1 } { 1 } \left ( x \sin\left( x \right) - \left ( \left [ \displaystyle\int { \color{#FF6800}{ - } \color{#FF6800}{ 1 } } d { \color{#FF6800}{ u } } \right ] _ { \color{#FF6800}{ u } = \color{#FF6800}{ \cos\left( \color{#FF6800}{ x } \right) } } \right ) + \left ( 1 - 1 \right ) \displaystyle\int { x \cos ^ { 1 - 2 } \left ( x \right) } d { x } \right )$
$\dfrac { 1 } { 1 } \left ( x \sin\left( x \right) - \left ( \left [ \displaystyle\int { \color{#FF6800}{ - } \color{#FF6800}{ 1 } } d { \color{#FF6800}{ u } } \right ] _ { u = \cos\left( x \right) } \right ) + \left ( 1 - 1 \right ) \displaystyle\int { x \cos ^ { 1 - 2 } \left ( x \right) } d { x } \right )$
 It is $\int -f(x) dx = -\int f(x) dx$
$\dfrac { 1 } { 1 } \left ( x \sin\left( x \right) - \left ( \left [ \color{#FF6800}{ - } \left ( \displaystyle\int { \color{#FF6800}{ 1 } } d { \color{#FF6800}{ u } } \right ) \right ] _ { u = \cos\left( x \right) } \right ) + \left ( 1 - 1 \right ) \displaystyle\int { x \cos ^ { 1 - 2 } \left ( x \right) } d { x } \right )$
$\dfrac { 1 } { 1 } \left ( x \sin\left( x \right) - \left ( \left [ - \left ( \displaystyle\int { \color{#FF6800}{ 1 } } d { \color{#FF6800}{ u } } \right ) \right ] _ { u = \cos\left( x \right) } \right ) + \left ( 1 - 1 \right ) \displaystyle\int { x \cos ^ { 1 - 2 } \left ( x \right) } d { x } \right )$
 The indefinite integral of $1$ is $x$ . 
$\dfrac { 1 } { 1 } \left ( x \sin\left( x \right) - \left ( \left [ - \color{#FF6800}{ u } \right ] _ { u = \cos\left( x \right) } \right ) + \left ( 1 - 1 \right ) \displaystyle\int { x \cos ^ { 1 - 2 } \left ( x \right) } d { x } \right )$
$\dfrac { 1 } { 1 } \left ( x \sin\left( x \right) - \left ( \left [ \color{#FF6800}{ - } \color{#FF6800}{ u } \right ] _ { \color{#FF6800}{ u } = \color{#FF6800}{ \cos\left( \color{#FF6800}{ x } \right) } } \right ) + \left ( 1 - 1 \right ) \displaystyle\int { x \cos ^ { 1 - 2 } \left ( x \right) } d { x } \right )$
 Return the substituted value 
$\dfrac { 1 } { 1 } \left ( x \sin\left( x \right) - \left ( \color{#FF6800}{ - } \color{#FF6800}{ \cos\left( \color{#FF6800}{ x } \right) } \right ) + \left ( 1 - 1 \right ) \displaystyle\int { x \cos ^ { 1 - 2 } \left ( x \right) } d { x } \right )$
$\dfrac { 1 } { 1 } \left ( x \sin\left( x \right) - \left ( - \cos\left( x \right) \right ) + \left ( \color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) \displaystyle\int { x \cos ^ { 1 - 2 } \left ( x \right) } d { x } \right )$
 Add $1$ and $- 1$
$\dfrac { 1 } { 1 } \left ( x \sin\left( x \right) - \left ( - \cos\left( x \right) \right ) + \color{#FF6800}{ 0 } \displaystyle\int { x \cos ^ { 1 - 2 } \left ( x \right) } d { x } \right )$
$\dfrac { 1 } { 1 } \left ( x \sin\left( x \right) - \left ( - \cos\left( x \right) \right ) + \color{#FF6800}{ 0 } \displaystyle\int { x \cos ^ { 1 - 2 } \left ( x \right) } d { x } \right )$
 If you multiply a number by 0, it becomes 0 
$\dfrac { 1 } { 1 } \left ( x \sin\left( x \right) - \left ( - \cos\left( x \right) \right ) + \color{#FF6800}{ 0 } \right )$
$\dfrac { 1 } { 1 } \left ( x \sin\left( x \right) - \left ( - \cos\left( x \right) \right ) \color{#FF6800}{ + } \color{#FF6800}{ 0 } \right )$
 0 does not change when you add or subtract 
$\dfrac { 1 } { 1 } \left ( x \sin\left( x \right) - \left ( - \cos\left( x \right) \right ) \right )$
$\color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 1 } } } \left ( x \sin\left( x \right) - \left ( - \cos\left( x \right) \right ) \right )$
 Calculate the value 
$\color{#FF6800}{ 1 } \left ( x \sin\left( x \right) - \left ( - \cos\left( x \right) \right ) \right )$
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