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∫sin4xcos3x dx

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Question

`int sin4x cos3x  "d"x`

Sum
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Solution

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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Chapter 2.3: Indefinite Integration - Short Answers I

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