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int0^pilog(1+cosx)dx ....

`int_0^pilog(1+cosx)dx` .

A

` - ( pi ) /( 2 ) log 2 `

B

` pi log "" ( 1 ) /( 2 ) `

C

` pi log 2 `

D

` ( pi ) /( 2 ) log 2 `

Text Solution

Verified by Experts

The correct Answer is:
B
Let I ` = int _ 0 ^ ( pi) log ( 1 + cos x ) dx " " `… (i)
I = ` int _ 0 ^( pi ) log { 1 + cos (pi- x )} dx `
` = int_ 0 ^(pi ) log ( 1 - cos x ) dx " " `… (ii)
On adding Eqs. (i) and (ii), we get
` 2I = int _ 0 ^( pi ) {log ( 1 + cos x ) + log ( 1 - cos x ) } dx `
` I = ( 1 ) /( 2 ) int _ 0 ^( pi) log ( 1 - cos ^( 2 ) x ) dx `
` = ( 1 ) /( 2 ) int _0 ^(pi) log sin ^2 x dx `
` = int _ 0^(pi ) log sin x dx `
` = 2 int _ 0 ^(pi // 2 ) log sin x dx `
` { because int _0 ^ ( 2a ) f ( x ) dx = 2 int_ 0 ^ ( a ) f (x ) dx, if f ( 2a - x ) =f ( x ) } `
` = 2{ - ( pi ) / ( 2 ) log 2} `
` ( because int _ 0 ^( pi//2 ) log sin x dx = - ( pi ) /( 2 ) log 2 ) `
` = pi log"" ( 1 )/ ( 2 ) `
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