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प्रश्न
By using the properties of the definite integral, evaluate the integral:
`int_0^pi log(1+ cos x) dx`
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उत्तर
Let `I = int_0^pi log (1 + cos x) dx` ....(i)
`I = int_0^pi log [1 + cos (pi - x)] dx`
`[∵ int_0^a f (x) dx = int_0^a f (a - x) dx]`
`= int_0^pi log (1 - cos x) dx` .....(ii)
Adding (i) and (ii), we get
`2 I = int_0^pi [log (1 + cos x) + log (1 - cos x)] dx`
`= int_0^pi log (1 - cos^2 x) dx`
`= int_0^pi log sin^2 x dx`
`= 2 int_0^pilog sin x dx`
⇒ `I = int_0^pi log sin x dx`
`= 2 int_0^(pi/2) log sin x dx = 2I_1`
`[∵ int_0^(2a) f (x) dx = 2 int_0^a f (x) dx, "if" f (2a - x) = f (x)]`
Where `I_1 = int_0^(pi/2) log sin x dx` ...(iii)
Then, `I_1 = int_0^(pi/2) log sin (pi/2 - x) dx`
⇒ `I_1 = int_0^(pi/2) log cos x dx` ....(iv)
Adding (iii) and (iv), we get
`2I_1 = int_0^(pi/2) log sin x dx + int_0^(pi/2) log cos x dx`
`= int_0^(pi/2) (log sin x + log cos x) dx`
`= int_0^(pi/2) log (sin x cos x) dx`
`= int_0^(pi/2) log ((2sin x cos x)/2)`
`= int_0^(pi/2) log ((sin 2x)/2) dx`
`= int_0^(pi/2) log sin 2 x dx - int_0^(pi/2) log 2 dx`
`= int_0^(pi/2) log sin 2x dx - (log 2)[x]_0^(pi/2)`
`= int_0^(pi/2) log sin 2 x dx - (log 2) (pi/2 - 0)`
`= int_0^(pi/2) log sin 2x dx - pi/2 log 2`
`= I_2 - pi/2 log 2` .....(v)
Where `I_2 = int_0^(pi/2) log sin 2x dx`
Put 2x = t
⇒ 2dx = dt
When x = 0, t = 0
When `x = pi/2, t = pi`
∴ `I_2 = 1/2 int_0^pi log sin t dt`
`= 1/2 *2 int_0^(pi/2) log sin t dt` `...[∵ log sin (pi -t) = log sint]`
`= int_0^(pi/2) log sin x = I_1`
∴ From (v), we get
`2I_1 = I_2 - pi/2 log 2`
⇒ `2I_1 = I_1 - pi/2 log 2`
⇒ `I_1 = pi/2 log 2`
∴ `I = 2 xx (-pi/2 log 2)`
= - π log 2
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