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प्रश्न
`int x sin2x cos5x "d"x`
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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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