Mathematics

# Solve:$\int x^2\cos^3x.dx$

##### SOLUTION
Let, $I=\int { { x }^{ 2 }{ cos }^{ 3 }x } dx$
We know the integration by parts formula,
$\int { u{ v }^{ ' } } dx=uv-\int { v{ u }^{ ' } } dx$
For $I=\int { { x }^{ 2 }{ cos }^{ 3 }x } dx$,
$u={ x }^{ 2 }\Rightarrow { u }^{ ' }=2x\\ { v }^{ ' }=\cos { 3x } \Rightarrow v=\dfrac { 1 }{ 3 } \sin { 3x } \\ \therefore I=\dfrac { { x }^{ 2 } }{ 3 } \sin { 3x } -\int { \left[ \dfrac { 2x }{ 3 } \sin { 3x } \right] } dx\\ =\dfrac { { x }^{ 2 } }{ 3 } \sin { 3x } -\dfrac { 2 }{ 3 } \int { \left[ x\sin { 3x } \right] } dx$
Let, ${ I }_{ 1 }=\int { \left( x\sin { 3x } \right) } dx$
Using integration by parts for ${ I }_{ 1 }=\int { \left( x\sin { 3x } \right) } dx$,
$u=x\Rightarrow { u }^{ ' }=1\\ { v }^{ ' }=\sin { 3x } \Rightarrow v=-\dfrac { 1 }{ 3 } \cos { 3x } \\ \therefore { I }_{ 1 }=(-\dfrac { x }{ 3 } \cos { 3x } )-[\int { (-\dfrac { 1 }{ 3 } \cos { 3x } ) } dx]\\ =-\dfrac { x }{ 3 } \cos { 3x } +\dfrac { 1 }{ 9 } \sin { 3x } \\ \therefore I=\dfrac { { x }^{ 2 } }{ 3 } \sin { 3x } -\dfrac { 2 }{ 3 } (-\dfrac { x }{ 3 } \cos { 3x } +\dfrac { 1 }{ 9 } \sin { 3x } )+C\\ =\dfrac { { x }^{ 2 } }{ 3 } \sin { 3x } +\dfrac { 2 }{ 9 } x\cos { 3x } -\dfrac { 2 }{ 27 } \sin { 3x } +C$
Hence, $\int { { x }^{ 2 }{ cos }^{ 3 }x } dx=\dfrac { { x }^{ 2 } }{ 3 } \sin { 3x } +\dfrac { 2 }{ 9 } x\cos { 3x } -\dfrac { 2 }{ 27 } \sin { 3x } +C.$

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

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