Mathematics

$$\displaystyle \int _{ 0 }^{ \pi  }{ x\sin { 2x } .dx } $$


SOLUTION
$$\int _{ 0 }^{ \Pi  }{ x\sin { 2x } .dx } $$
Apply bypart rule
$$\int _{ 0 }^{ \Pi  }{ x\sin { 2x } .dx } \\ =-\frac { 1 }{ 2 } x\cos { 2x } -\int { -\frac { 1 }{ 2 } \cos { 2x\quad dx }  } \\ =-\frac { 1 }{ 2 } x\cos { 2x\quad +\quad \frac { 1 }{ 4 }  } \sin { 2x } $$
Now compute the boundary conditions
$$={ \left[ -\frac { 1 }{ 2 } x\cos { 2x\quad +\quad \frac { 1 }{ 4 }  } \sin { 2x } \\  \right]  }_{ 0 }^{ \Pi  }\\ =-\frac { \Pi  }{ 2 } -0\\ =-\frac { \Pi  }{ 2 }$$


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Subjective Medium Published on 17th 09, 2020
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