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Question
Integrate the functions:
`1/(1 + cot x)`
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Solution
Let `I = int 1/ (1 + cot x) dx = int 1/ (1 + cos x/sinx) dx`
`= int sin/(sin x + cos x) dx`
`= 1/2 int (2 sin x)/ (sinx + cos x) dx`
`= 1/2 int ((sin x + cos x) - (cos x - sin x))/ ((sin x + cos x)) dx`
`= 1/2 int 1 dx - 1/2 int (cos x - sin x)/ (sin x + cos x) dx`
`= 1/2 x - 1/2 int (cos x - sin x)/ (sin x + cos x) dx + C_1`
`I = x/2 - 1/2 I_1 + C_1` ........(i)
Where, `I_1 = int (cos x - sin x)/ (sin x + cos x) dx`
Put sin x + cos x = t
⇒ (cos x - sin x) dx = dt
⇒ `I_1 = int dt/t = log |t| + C_2`
`= log |cos x + sin x| + C_2` ......(ii)
From (i) and (ii), we get
⇒ `I = 1/2 x - 1/2 log |cos x + sin x| + C`
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