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Question
Integrate the following with respect to x:
x sin 3x
Solution
`int x sin 3x "d"x`
`int "u" "dv" = "uv" - "u""'""v"_1 + "u""'""v"_2 - "u""'""v"_3 +` ..........(1)
u = x
u' = 1
u" = 0
v1 = `int "v" "d"x``
= `int - (cos 3x)/3`
= `- 1/3 xx (sin 3x)/3`
v2 = `int "v"_1 "d"x`
= `int - 1/3^2 sin3x * "d"x`
= `- 1/3^2 xx 1/3 xx - cos 3x`
= `1/3^2 cos 3x`
`int x sin x "d"x = x xx - (cos3x)/3 - 1 xx - 1/3^2 sin 3x + 0 xx 1/3^3 cos3x + "c"`
`int x sin 3x "d"x = - x/3 cos 3x + 1/9 sin 3x + "c"`
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