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Evaluate the following : ∫x.sin2x.dx

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Question

Evaluate the following : `int x.sin^2x.dx`

Sum
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Solution

`int x.sin^2x.dx`

= `int x((1 - cos2x)/2).dx`

= `(1)/(2) int (x - x cos2x).dx`

= `(1)/(2) int x.dx - (1)/(2) int x cos 2x.dx`

= `(1)/(2).x^2/(2) - (1)/(2)[x int cos 2x.dx - int {d/dx (x) int cos 2x.dx}.dx]`

= `x^2/(4) - (1)/(2)[x. (sin2x)/(2) - int 1. (sin2x)/(2).dx]`

= `x^2/(4) - (1)/(2) x. sin2x + (1)/(4) sin 2x.dx`

= `x^2/(4) - (1)/(4) x.sin2x + (1)/(4).((-cos2x))/(2) + c`

= `x^2/(4) - (1)/(4) x.sin2x - (1)/(8) cos 2x + c`

= `(1)/(4) [x^2 - x.sin 2x - (1)/(2) cos 2x] + c`.

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Chapter 3: Indefinite Integration - Exercise 3.3 [Page 137]

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