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Integrals

Question
CBSEENMA12032609

Show that:
integral subscript 0 superscript 1 fraction numerator log space left parenthesis 1 plus straight x right parenthesis over denominator 1 plus straight x squared end fraction space equals space straight pi over 8 space log space 2

Solution

Let I = integral subscript 0 superscript 1 fraction numerator log left parenthesis 1 plus straight x right parenthesis over denominator 1 plus straight x squared end fraction dx
Put x = tan θ,    ∴ dx = sec2 θ dθ
When x = 0, tan θ = 0 ⇒ θ = 0
When x = 1,  tan space straight theta space equals space 1 space space space space space space space space rightwards double arrow space space space space straight theta space equals space straight pi over 4

therefore space space space straight I space equals space integral subscript 0 superscript straight pi over 4 end superscript space log left parenthesis 1 plus tanθ right parenthesis space dθ space                                         ...(1)
           equals space integral subscript 0 superscript straight pi over 4 end superscript log space open square brackets 1 plus tan open parentheses straight pi over 4 minus straight theta close parentheses close square brackets space dθ   open square brackets because space integral subscript 0 superscript straight a straight f left parenthesis straight x right parenthesis space dx space equals space integral subscript 0 superscript straight a straight f left parenthesis straight a minus straight x right parenthesis space dx close square brackets
           equals space integral subscript 0 superscript straight pi over 4 end superscript log open square brackets 1 plus fraction numerator 1 minus tanθ over denominator 1 plus tanθ end fraction close square brackets dθ space equals space integral subscript 0 superscript straight pi over 4 end superscript log open square brackets fraction numerator 1 plus tanθ plus 1 minus tanθ over denominator 1 plus tanθ end fraction close square brackets dθ space equals space integral subscript 0 superscript straight pi over 4 end superscript log open parentheses fraction numerator 2 over denominator 1 plus tanθ end fraction close parentheses dθ
 equals space integral subscript 0 superscript straight pi over 4 end superscript log space 2 space dθ space space minus integral subscript 0 superscript straight pi over 4 end superscript log left parenthesis 1 plus tanθ right parenthesis space dθ
therefore space space space space straight I space equals space log space 2 integral subscript 0 superscript straight pi over 4 end superscript space 1. space dθ space minus space 1                                                          [because of (1)]
rightwards double arrow space 2 space straight I space space equals space log space 2 open square brackets straight theta close square brackets subscript 0 superscript straight pi over 4 end superscript space equals space open parentheses straight pi over 4 minus 0 close parentheses space log space 2 space rightwards double arrow space space 2 space straight I space equals space straight pi over 4 log 2 space space rightwards double arrow space space straight I equals straight pi over 8 log space 2

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