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Question
Evaluate the following:
`int ((cos 5x + cos 4x))/(1 - 2 cos 3x) "d"x`
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Solution
Let I = `int ((cos 5x + cos 4x))/(1 - 2 cos 3x) "d"x`
= `int (2cos (5x + 4x)/2 * cos (5x - 4x)/2)/(1 - 2(2 cos^2 (3x)/2 - 1)) "d"x`
= `int (2cos (9x)/2 * cos x/2)/(1 - 4 cos^2 (3x)/2 + 2) "d"x`
= `int (2cos (9x)/2 * cos x/2)/(3 - 4 cos^2 (3x)/2) "d"x`
= `- int (2 cos (9x)/2 * cos x/2)/(4 cos^2 (3x)/2 - 3) "d"x`
= `- int (2cos (9x)/2 * cos x/2 * cos (3x)/2)/(4 cos^2 (3x)/2 - 3 cos (3x)/2) "d"x` ....`["Multiplying and dividing by" cos (3x)/2]`
= `int (2 cos (9x)/2 * cos x/2 * cos (3x)/2)/(cos 3 * (3x)/2) "dx"` ......[∵ cos 3x = 4 cos3x – 3 cos x]
= `- int (2cos (9x)/2 * cos x/2 * cos (3x)/2)/(cos (9x)/2) "d"x`
= `- int 2 cos (3x)/2 * cos x/2 "d"x`
= `- int [cos((3x)/2 + x/2) + cos((3x)/2 - x/2)] "d"x`
= `- int (cos 2x + cos x) "d"x` ....[∵ 2 cos A cos B = cos (A + B) + cos (A – B)]
= `- int cos 2x "d"x - int cos x "d"x`
= `- 1/2 sin 2x - sin x + "C"`
Hence, I = `- [1/2 sin 2x + sin x] + "C"`.
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