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∫xsin2xcos5x dx

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प्रश्न

`int x sin2x cos5x  "d"x`

योग
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उत्तर

Let I = `int x sin2x cos5x  "d"x`

= `1/2 int x (2 sin 2x cos 5x)  "d"x`

= `1/2 int  x [sin (2x + 5x) + sin(2x - 5x)]  "d"x`

= `1/2 int x [sin 7x - sin (-3x)]  "d"x` 

= `1/2 int x (sin 7x - sin 3x)  "d"x`

= `1/2 int x sin 7x  "d"x - 1/2 int  x sin 3x  "d"x`

= `1/2 [x int sin 7x  "d"x - int {"d"/("d"x) (x) int sin 7x  "d"x}"d"x] - 1/2 [x int sin 3x  "d"x - int {"d"/("d"x)(x) int sin 3x  "d"x}"d"x]`

= `1/2[x(- (cos 7x)/7) - int 1* ((-cos 7x)/7) "d"x] - 1/2[x((-cos 3x)/3) - int 1* ((-cos 3x)/3) "d"x]`

= `1/2 ((-x cos 7x)/7 + 1/7 int cos 7x  "d"x) - 1/2((-x cos 3x)/3 + 1/3 int cos 3x  "d"x)`

= `1/2[(-x cos 7x)/7 + 1/7((sin7x)/7)] - 1/2[(-x cos 3x)/3 + 1/3((sin 3x)/3)] + "c"`

∴ I = `1/98  sin 7x - 1/14  x cos 7x - 1/18  sin 3x + 1/6  x cos 3x + "c"`

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अध्याय 2.3: Indefinite Integration - Long Answers III

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