Mathematics

# Integrate : $\int x\sin^2x$

##### SOLUTION
$\int { x{ \sin }^{ 2 }xdx } \\ =\int { x\left[ \dfrac { 1 }{ 2 } (1-\cos2x) \right] dx } \quad (\because \cos2x=1-2{ \sin }^{ 2 }x)\\ =\dfrac { 1 }{ 2 } \int { xdx } -\dfrac { 1 }{ 2 } \int { x\cos2xdx } \\ =\dfrac { { x }^{ 2 } }{ 4 } -\dfrac { 1 }{ 2 } \left[ x\int { \cos2xdx } -\int { \left( \dfrac { dx }{ dx } .\int { \cos2xdx } \right) } \right] dx\\ =\dfrac { { x }^{ 2 } }{ 4 } -\dfrac { x\sin2x }{ 4 } -\dfrac { \cos2x }{ 8 } +C\\ =\dfrac { { 2x }^{ 2 }-2x\sin2x-\cos2x }{ 8 } +C$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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