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Question
Evaluate the following:
`int x^2 sin 3x dx`
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Solution
Let I = `int x^2 sin 3x dx`
= `x^2 int sin 3x.dx - int [d/dx (x^2) int sin 3x.dx]dx ...[∵ int uv.dx = uintv.dx - int[(du)/(dx) int v.dx]dx]`
= `x^2(-(cos3x)/3) - int2x(-(cos3x)/3).dx`
= `-x^2/3 cos3x + (2)/(3) int x cos 3x dx`
= `-x^2/3 cos3x + (2)/(3)[x int cos 3x dx - int {d/dx (x) int cos 3x .dx} .dx] ...[∵ int uv.dx = uintv.dx - int[(du)/(dx) int v.dx]dx]`
= `-x^2/3 cos3x + 2/3[(xsin3x)/(3) - int 1. (sin3x)/(3).dx]`
= `-x^2/3 cos3x + (2 x sin 3x)/9 - (2)/(9) int (sin 3x)/3 dx`
= `-x^2/3 cos3x + (2 x sin 3x)/9 - (2)/(9) ((- cos3x)/3) + c`
= `-x^2/3 cos3x + (2 x sin 3x)/9 + (2 cos 3x)/27 + c`
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