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प्रश्न
`int sin4x cos3x "d"x`
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उत्तर
Let I = `int sin 4x * cos3x "d"x`
= `1/2 int (2 sin 4x * cos 3x) "d"x`
= `1/2 int [sin (4x + 3x) + sin(4x - 3x)] "d"x` .......[∵ 2 sin A cos B = sin(A + B) + sin(A − B)]
= `1/2 int (sin 7x + sin x) "d"x`
= `1/2 [int sin7 x "d"x + int sin x "d"x]`
= `1/2((-cos7x)/7 - cos x) + "c"`
∴ I = `- 1/14 cos 7x - 1/2 cos x + "c"`
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