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Question
Integrate the following with respect to x:
x3 sin x
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Solution
`int x^3 sin x`
u = x3
u’ = 3x2
u” = 6x
u”’ = 6
u’v = 0
dv = sin x dx
⇒ v = `int sin x "d"x`
= – cosx
v1 = `int "v" "d"x`
= `int - cos x "d"x`
= `- int cos x "d"x`
= – sin x
v2 = `int "v"_1 "d"x`
= `int - sin x "d"x`
= `- int sin x "d"x`
= – (– cos x)
= cos x
v3 = `int "v"_2 "d"x`
= `int cos x "d"x`
= sin x
v4 = `int "v"_3 "d"x`
= `int sin x "d"x`
= – cos x
`int "u" "dv"` = uv – u’ v1 + u” v2 – u”’ v3 + u’vv4 – ………..
`int x^3 sin x "d"x = x^3 (- cos x) – 3x^2 (- sin x) + 6x (cos x) – 6 sin x + 0 (- cos x) + "c"`
= – x3 cos x + 3x2 sin x + 6x cos x – 6 sin x + c
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