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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 111

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