anonymous
  • anonymous
xsin(3x)dx @LeibyStrauss
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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anonymous
  • anonymous
\[\int\limits_{}^{}udv = uv - \int\limits_{}^{}vdu\]u = x du = 1 dx dv = sin3x To find v, let's redo \[\int\limits_{}^{}\sin(3x)\]
anonymous
  • anonymous
\[\int\limits_{}^{}\sin 3x\] u = 3x du = 3 dx we want some du should equal 1 dx so we divide both sides by 3 1/3 du = dx \[\frac{ 1 }{ 3 } \int\limits_{}^{}\sin(u) du => -\frac{ 1 }{ 3 } \cos(u) => -\frac{ 1 }{ 3 } \cos(3x)\] v = -1/3 cos(3x)
anonymous
  • anonymous
\[v = 1/3-\cos(3x)\]

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anonymous
  • anonymous
so let's plug the values in
anonymous
  • anonymous
\[= uv - \int\limits_{}^{}vdu \] \[= x [-\frac{ 1 }{ 3 }\cos(3x)] - \int\limits_{}^{}-\frac{ 1 }{ 3 }\cos(3x) dx\] so we need to integrate \[\int\limits_{}^{}-\frac{ 1 }{ 3 }\cos(3x) dx\]
anonymous
  • anonymous
\[1/3\int\limits_{ }^{} \cos(3x)\]
anonymous
  • anonymous
intergration of cos(3x)
anonymous
  • anonymous
\[\int\limits_{}^{}-\frac{ 1 }{ 3 }\cos(3x) dx => -\frac{ 1 }{ 3 }\int\limits_{}^{}\cos(3x) dx\] u = 3x du = 3dx 1/3 du = dx \[-\frac{ 1 }{ 3 } \int\limits_{}^{}\cos(u) \frac{ 1 }{ 3 }du => -\frac{ 1 }{ 3 }\frac{ 1 }{ 3 } \int\limits_{}^{}\cos(u) du\] \[= -\frac{ 1 }{ 9 }\sin(3x) + C\]
anonymous
  • anonymous
in the very 1st step dv = sin(3x) d/dx of cos = -sin, d/dx of -cos = sin \[\int\limits_{}^{}\sin(x) = -\cos(x)\]
anonymous
  • anonymous
\[x[-\frac{ 1 }{ 3 }\cos(3x)] + \frac{ 1 }{ 9 }\sin(3x) + C => -\frac{ x }{ 3 }\cos(3x) + \frac{ 1 }{ 9 }\sin(3x) + C\]

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