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Question
Evaluate the following:
`int x.sin 2x. cos 5x.dx`
Solution
Let I = `int x.sin 2x. cos 5x.dx`
`sin 2x cos 5x = (1)/(2)[2 sin2x cos5x]`
= `(1)/(2)[sin(2x+ 5x) + sin(2x - 5x)]`
= `(1)/(2)[sin7x - sin3x]`
∴ `int sin 2x cos 5x .dx = (1)/(2)[int sin 7x ..dx - intsin 3x.dx]`
= `(1)/(2)((-cos7x)/7) - (1)/(2) ((- cos3x)/3)`
= `-(1)/(14) cos7x + (1)/(6) cos3x` ...(1)
I = `int x sin 2x cos 5x.dx`
= `x int sin 2x cos 5x.dx - int [d/dx (x) int sin 2x cos 5x.dx].dx`
= `x[-1/14 cos7x + 1/6 cos 3x] - int 1.(-1/14 cos7x + 1/6 cos3x).dx` ...[By (1)]
= `-x/(14) cos7x + x/(6) cos3x + (1)/(14) int cos7x.dx - (1)/(6) int cos 3x.dx`
= `-x/(14) cos7x + x/(6) cos3x + (1)/(14) ((sin7x)/7) - (1)/(6) ((sin3x)/3) + c`
= `- x/(14) cos7x + x/(6) cos3x + (sin7x)/(98) - (sin3x)/(18) + c`.
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