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Calculate the integral
Answer
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$\int{ x \sin\left( x \right) }d{ x }$
$1 \left ( - x \cos\left( x \right) + \sin\left( x \right) \right )$
Calculate the integral
$\displaystyle\int { \color{#FF6800}{ x } \color{#FF6800}{ \sin\left( x \right) } } d { \color{#FF6800}{ x } }$
$ $ Calculate the integral using the formula of $ \int x \sin^{n}(x) dx =\dfrac{1}{n}(-x\cos(x)\sin^{n-1}(x)+\int{\cos(x)\sin^{n-1}(x)}d{x}+(n-1)\int{x\sin^{n-2}(x)}d{x})$
$\color{#FF6800}{ \dfrac { 1 } { 1 } } \left ( \color{#FF6800}{ - } \color{#FF6800}{ x } \color{#FF6800}{ \cos\left( x \right) } \color{#FF6800}{ \sin ^ { 1 - 1 } \left ( x \right) } \color{#FF6800}{ + } \displaystyle\int { \color{#FF6800}{ \cos\left( x \right) } \color{#FF6800}{ \sin ^ { 1 - 1 } \left ( x \right) } } d { \color{#FF6800}{ x } } \color{#FF6800}{ + } \left ( \color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) \displaystyle\int { \color{#FF6800}{ x } \color{#FF6800}{ \sin ^ { 1 - 2 } \left ( x \right) } } d { \color{#FF6800}{ x } } \right )$
$\dfrac { 1 } { 1 } \left ( - x \cos\left( x \right) \sin ^ { \color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } } \left ( x \right) + \displaystyle\int { \cos\left( x \right) \sin ^ { 1 - 1 } \left ( x \right) } d { x } + \left ( 1 - 1 \right ) \displaystyle\int { x \sin ^ { 1 - 2 } \left ( x \right) } d { x } \right )$
$ $ Add $ 1 $ and $ - 1$
$\dfrac { 1 } { 1 } \left ( - x \cos\left( x \right) \sin ^ { \color{#FF6800}{ 0 } } \left ( x \right) + \displaystyle\int { \cos\left( x \right) \sin ^ { 1 - 1 } \left ( x \right) } d { x } + \left ( 1 - 1 \right ) \displaystyle\int { x \sin ^ { 1 - 2 } \left ( x \right) } d { x } \right )$
$\dfrac { 1 } { 1 } \left ( - x \cos\left( x \right) \color{#FF6800}{ \sin ^ { 0 } \left ( x \right) } + \displaystyle\int { \cos\left( x \right) \sin ^ { 1 - 1 } \left ( x \right) } d { x } + \left ( 1 - 1 \right ) \displaystyle\int { x \sin ^ { 1 - 2 } \left ( x \right) } d { x } \right )$
$ $ Calculate power $ $
$\dfrac { 1 } { 1 } \left ( - x \cos\left( x \right) \times \color{#FF6800}{ 1 } + \displaystyle\int { \cos\left( x \right) \sin ^ { 1 - 1 } \left ( x \right) } d { x } + \left ( 1 - 1 \right ) \displaystyle\int { x \sin ^ { 1 - 2 } \left ( x \right) } d { x } \right )$
$\dfrac { 1 } { 1 } \left ( \color{#FF6800}{ - } \color{#FF6800}{ x } \color{#FF6800}{ \cos\left( x \right) } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } + \displaystyle\int { \cos\left( x \right) \sin ^ { 1 - 1 } \left ( x \right) } d { x } + \left ( 1 - 1 \right ) \displaystyle\int { x \sin ^ { 1 - 2 } \left ( x \right) } d { x } \right )$
$ $ Multiplying any number by 1 does not change the value $ $
$\dfrac { 1 } { 1 } \left ( \color{#FF6800}{ - } \color{#FF6800}{ x } \color{#FF6800}{ \cos\left( x \right) } + \displaystyle\int { \cos\left( x \right) \sin ^ { 1 - 1 } \left ( x \right) } d { x } + \left ( 1 - 1 \right ) \displaystyle\int { x \sin ^ { 1 - 2 } \left ( x \right) } d { x } \right )$
$\dfrac { 1 } { 1 } \left ( - x \cos\left( x \right) + \displaystyle\int { \cos\left( x \right) \sin ^ { \color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } } \left ( x \right) } d { x } + \left ( 1 - 1 \right ) \displaystyle\int { x \sin ^ { 1 - 2 } \left ( x \right) } d { x } \right )$
$ $ Add $ 1 $ and $ - 1$
$\dfrac { 1 } { 1 } \left ( - x \cos\left( x \right) + \displaystyle\int { \cos\left( x \right) \sin ^ { \color{#FF6800}{ 0 } } \left ( x \right) } d { x } + \left ( 1 - 1 \right ) \displaystyle\int { x \sin ^ { 1 - 2 } \left ( x \right) } d { x } \right )$
$\dfrac { 1 } { 1 } \left ( - x \cos\left( x \right) + \displaystyle\int { \color{#FF6800}{ \cos\left( x \right) } \color{#FF6800}{ \sin ^ { 0 } \left ( x \right) } } d { \color{#FF6800}{ x } } + \left ( 1 - 1 \right ) \displaystyle\int { x \sin ^ { 1 - 2 } \left ( x \right) } d { x } \right )$
$ $ Substitute with $ u = \sin\left( x \right) $ and calculate the integral $ $
$\dfrac { 1 } { 1 } \left ( - x \cos\left( x \right) + \left [ \displaystyle\int { \color{#FF6800}{ 1 } } d { \color{#FF6800}{ u } } \right ] _ { \color{#FF6800}{ u } = \color{#FF6800}{ \sin\left( x \right) } } + \left ( 1 - 1 \right ) \displaystyle\int { x \sin ^ { 1 - 2 } \left ( x \right) } d { x } \right )$
$\dfrac { 1 } { 1 } \left ( - x \cos\left( x \right) + \left [ \displaystyle\int { \color{#FF6800}{ 1 } } d { \color{#FF6800}{ u } } \right ] _ { u = \sin\left( x \right) } + \left ( 1 - 1 \right ) \displaystyle\int { x \sin ^ { 1 - 2 } \left ( x \right) } d { x } \right )$
$ $ The indefinite integral of $ 1 $ is $ x $ . $ $
$\dfrac { 1 } { 1 } \left ( - x \cos\left( x \right) + \left [ \color{#FF6800}{ u } \right ] _ { u = \sin\left( x \right) } + \left ( 1 - 1 \right ) \displaystyle\int { x \sin ^ { 1 - 2 } \left ( x \right) } d { x } \right )$
$\dfrac { 1 } { 1 } \left ( - x \cos\left( x \right) + \left [ \color{#FF6800}{ u } \right ] _ { \color{#FF6800}{ u } = \color{#FF6800}{ \sin\left( x \right) } } + \left ( 1 - 1 \right ) \displaystyle\int { x \sin ^ { 1 - 2 } \left ( x \right) } d { x } \right )$
$ $ Return the substituted value $ $
$\dfrac { 1 } { 1 } \left ( - x \cos\left( x \right) + \color{#FF6800}{ \sin\left( x \right) } + \left ( 1 - 1 \right ) \displaystyle\int { x \sin ^ { 1 - 2 } \left ( x \right) } d { x } \right )$
$\dfrac { 1 } { 1 } \left ( - x \cos\left( x \right) + \sin\left( x \right) + \left ( \color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) \displaystyle\int { x \sin ^ { 1 - 2 } \left ( x \right) } d { x } \right )$
$ $ Add $ 1 $ and $ - 1$
$\dfrac { 1 } { 1 } \left ( - x \cos\left( x \right) + \sin\left( x \right) + \color{#FF6800}{ 0 } \displaystyle\int { x \sin ^ { 1 - 2 } \left ( x \right) } d { x } \right )$
$\dfrac { 1 } { 1 } \left ( - x \cos\left( x \right) + \sin\left( x \right) + \color{#FF6800}{ 0 } \displaystyle\int { x \sin ^ { 1 - 2 } \left ( x \right) } d { x } \right )$
$ $ If you multiply a number by 0, it becomes 0 $ $
$\dfrac { 1 } { 1 } \left ( - x \cos\left( x \right) + \sin\left( x \right) + \color{#FF6800}{ 0 } \right )$
$\dfrac { 1 } { 1 } \left ( - x \cos\left( x \right) + \sin\left( x \right) \color{#FF6800}{ + } \color{#FF6800}{ 0 } \right )$
$ $ 0 does not change when you add or subtract $ $
$\dfrac { 1 } { 1 } \left ( - x \cos\left( x \right) + \sin\left( x \right) \right )$
$\color{#FF6800}{ \dfrac { 1 } { 1 } } \left ( - x \cos\left( x \right) + \sin\left( x \right) \right )$
$ $ Calculate the value $ $
$\color{#FF6800}{ 1 } \left ( - x \cos\left( x \right) + \sin\left( x \right) \right )$
Solution search results
search-thumbnail-If the sum of two consecutive 
numbers is $45$ and one number is $X$ 
.This statement in the form of 
equation $1s:$ 
$\left(1$ Point) $\right)$ 
$○5x+1$ $1eft\left(x+1$ $r1gnt\right)=45s$ 
$○sx+1ef\left(x+2$ $r1gnt\right)=145s$ 
$sx+1x=45s$
7th-9th grade
Algebra
search-thumbnail-$s|ef\left(-1n$ $\left($ }\right)^{50}\ $\right)$ \ | | is\ equal\ to\ $S$ 
$s1S$ 
$S-1S$ 
$s2S$ 
$s50s$
7th-9th grade
Other
search-thumbnail-Given the set of ordered pairs $\left(\left(-7.0\right),\left(-6,5\right),\left(-5,-3\right),\left(-1,2\right)$ $\left(1,6\right),\left(2,-2\right)$ $\left(5,3\right)\left(7,-8\right)\right)$ 
Find f(7)fAleft(7\right) 
O a 
O b -8 
6. 
$5$
7th-9th grade
Algebra
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