InterviewSolution
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InterviewSolution
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`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 ` |
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Answer» Correct Answer - 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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