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Question
Find the integrals of the function:
cos 2x cos 4x cos 6x
Solution
Let `I = int cos 2x cos 4x cos 6x` dx
`= 1/2 int (2 cos 2x cos 4x) cos 6x dx`
... [∵ 2 cos A cos B = cos (A + B) - cos (A - B)]
`= 1/2 int (cos 6x + cos 2x) cos 6x dx`
`= 1/4 int 2 cos^2 6x dx + 1/4 int (2 cos 2c cos 6x) dx`
`= 1/4 int (1 + cos 12 x) dx + 1/4 int (cos 8x + cos 4x)` dx
`= 1/4 x + 1/4 ((sin 12x)/12) + 1/4 ((sin 8x)/8 + (sin 4x)/4) + C`
`= 1/4 [x + 1/12 sin 12 x + 1/8 sin 8x + 1/4 sin 4x] + C`
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