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Question
Evaluate the following integrals : `int sin 4x cos 3x dx`
Solution
`int sin 4x cos 3x dx`
= `(1)/(2)int sin 4x cos 3x dx` ...[∴ 2sinA.cosB = sin (A + B) + sin(A - B)]
= `(1)/(2)int [sin (4x + 3x) + sin (4x - 3x)]dx`
= `(1)/(2) int sin 7x dx + (1)/(2)int sin x dx`
= `(-1)/(2)((cos 7x)/7) + ((-1)/(2))cos x + c`
= `-(1)/(14)cos 7x - (1)/(2) cos x + c`.
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