Question

xsin(3x)dx
@LeibyStrauss

Answer 1

\[\int\limits_{}^{}udv = uv - \int\limits_{}^{}vdu\]u = x
du = 1 dx dv = sin3x

To find v, let's redo

\[\int\limits_{}^{}\sin(3x)\]

Answer 2

\[\int\limits_{}^{}\sin 3x\]

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)

Answer 3

\[v = 1/3-\cos(3x)\]

Answer 4

so let's plug the values in

Answer 5

\[= uv - \int\limits_{}^{}vdu \]

\[= x [-\frac{ 1 }{ 3 }\cos(3x)] - \int\limits_{}^{}-\frac{ 1 }{ 3 }\cos(3x) dx\]

so we need to integrate

\[\int\limits_{}^{}-\frac{ 1 }{ 3 }\cos(3x) dx\]

Answer 6

\[1/3\int\limits_{ }^{} \cos(3x)\]

Answer 7

intergration of cos(3x)

Answer 8

\[\int\limits_{}^{}-\frac{ 1 }{ 3 }\cos(3x) dx => -\frac{ 1 }{ 3 }\int\limits_{}^{}\cos(3x) dx\]

u = 3x
du = 3dx
1/3 du = dx

\[-\frac{ 1 }{ 3 } \int\limits_{}^{}\cos(u) \frac{ 1 }{ 3 }du => -\frac{ 1 }{ 3 }\frac{ 1 }{ 3 } \int\limits_{}^{}\cos(u) du\]

\[= -\frac{ 1 }{ 9 }\sin(3x) + C\]

Answer 9

in the very 1st step dv = sin(3x)

d/dx of cos = -sin,

d/dx of -cos = sin

\[\int\limits_{}^{}\sin(x) = -\cos(x)\]

Answer 10

\[x[-\frac{ 1 }{ 3 }\cos(3x)] + \frac{ 1 }{ 9 }\sin(3x) + C

=>

-\frac{ x }{ 3 }\cos(3x) + \frac{ 1 }{ 9 }\sin(3x) + C\]