Mathematics

Solve $$\displaystyle  \int { { x }^{ 2 }.\sin {2 x }\ dx }$$


ANSWER

$$ \dfrac { {- x }^{ 2 }\cos { 2x } }{ 2 } +\dfrac { 2x\sin { 2x } }{ 4 } +\dfrac { 2\cos { 2x } }{ 8 } +c$$


SOLUTION
$$\int {{x^2}.\sin 2xdx} $$
$$ = {x^2} \times \dfrac{{ - \cos 2x}}{2} - 2x \times \dfrac{{ - \cos 2x}}{2}dx$$
$$ = \dfrac{{ - {x^2}\cos 2x}}{2} + \int {x\cos 2xdx} $$
$$ = \dfrac{{ - {x^2}\cos 2x}}{2} + x \times \dfrac{{\sin 2x}}{2} - \int {1 \times } \dfrac{{\sin 2x}}{2}dx$$
$$ = \dfrac{{ - {x^2}\cos 2x}}{2} + x \times \dfrac{{\sin 2x}}{2}-\dfrac{{ 1}}{2} \times \dfrac{{ - \cos 2x}}{2} + c$$
$$ = \dfrac{{ - {x^2}\cos 2x}}{2} + \dfrac{{x\sin 2x}}{2} + \dfrac{{\cos 2x}}{4} + c$$
$$ = \dfrac{{ - {x^2}\cos 2x}}{2} + \dfrac{{2x\sin 2x}}{4} + \dfrac{{2\cos 2x}}{8} + c$$
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Single Correct Medium Published on 17th 09, 2020
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