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Question
Evaluate the following:
`int (cosx - cos2x)/(1 - cosx) "d"x`
Solution
Let I = `int (cosx - cos2x)/(1 - cosx) "d"x`
= `int (2sin (x + 2x)/2 * sin ((2x - x)/2))/(2sin^2 x/2) "d"x` ......`[because cos "C" - cos "D" = 2 sin ("C" + "D")/2 * sin ("D" - "C")/2]`
= `int (2sin (3x)/2 * sin x/2)/(2sin^2 x/2) "d"x`
= `int (sin (3x)/2)/(sin x/2) "d"x`
= `int (sin 3(x/2))/(sin x/2) "d"x`
= `int (3 sin x/2 - 4 sin^3 x/2)/(sin x/2) "d"x` ....[sin 3x = 3 sin x – 4 sin3x]
= `int (sin x/2 (3 - 4 sin^2 x/2))/(sin x/2) "d"x`
= `int (3 - 4 sin^2 x/2) "d"x`
= `int [3 - 2(1 - cosx)]"d"x` ......`[because 2 sin^2 x/2 = 1 - cos x]`
= `int (3 - 2 + 2 cos x) "d"x`
= `int (1 + 2 cos x) "d"x`
= x + 2 sin x + C
Hence, I = x + 2 sin x + C.
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