Mathematics

$$\displaystyle \int x \sin \dfrac {x}{2}=?$$


SOLUTION
Sol:
$$\displaystyle \int x\sin \dfrac {x}{2}dx$$
Let
$$I=\displaystyle \int x\sin \dfrac {x}{2}dx$$
We know that
$$I=\displaystyle \int \Pi dx =I\displaystyle \int \Pi dx-\int \left (\dfrac {dI}{dx}\displaystyle \int \Pi dx\right)dx+C$$
Now, using $$LAT \varepsilon $$
$$I=\displaystyle \int { x\sin { \dfrac { x }{ 2 } dx } =x } \displaystyle \int { \sin { \dfrac { x }{ 2 } dx\quad - }  } \left[ \displaystyle \int { \left( \dfrac { dx }{ dx } \displaystyle \int { \sin { \dfrac { x }{ 2 }  } dx }  \right)  } dx \right] +C$$
$$=x\dfrac { \left[ -\cos { \dfrac { x }{ 2 }  }  \right]  }{ \dfrac { 1 }{ 2 }  } -\displaystyle \int { \dfrac { \left( -\cos { \dfrac { x }{ 2 }  }  \right)  }{ \dfrac { 1 }{ 2 }  }  } dx+C$$
$$=-2x\cos \dfrac {x}{2}+\displaystyle \int 2\cos \dfrac {x}{2}dx+C$$
$$=2x\cos \dfrac {x}{2}+2\displaystyle \int \cos \dfrac {x}{2}dx+C$$
$$=-2cos \dfrac {x}{2}+2\left [\dfrac {\dfrac {\sin x}{2}}{\dfrac {1}{2}}\right ]+C$$
$$=-cos \dfrac {x}{2}+4\sin \dfrac {x}{2}+C$$
Hence, this is the answer.

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