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Integrals

Question
CBSEENMA12032629

Show that:
integral subscript 0 superscript 1 log space open parentheses 1 over straight x minus 1 close parentheses dx space equals space 0

Solution

Let I = integral subscript 0 superscript 1 log space open parentheses 1 over straight x minus 1 close parentheses dx
Put x = cos2 θ, θ dx = 2 cos θ (– sin θ) dθ = – 2 sin θ cos θ dθ = – sin 2θ dθ
When x = 0,  cos squared straight theta space equals space 0 space rightwards double arrow space space space straight theta space equals space straight pi over 2
When x = 1, cos2 θ = 1 ⇒ θ = 0
therefore space space space straight I space equals space minus integral subscript straight pi over 2 end subscript superscript 0 log open parentheses fraction numerator 1 over denominator cos squared straight theta end fraction minus 1 close parentheses space sin space 2 straight theta space dθ space equals space integral subscript 0 superscript straight pi over 2 end superscript log space open parentheses fraction numerator 1 minus cos squared straight theta over denominator cos squared straight theta end fraction close parentheses space sin space 2 straight theta space dθ
                                                                                        open square brackets because space space space integral subscript straight a superscript straight b straight f left parenthesis straight x right parenthesis space dx space equals space minus integral subscript straight b superscript straight a straight f left parenthesis straight x right parenthesis space dx close square brackets
                equals space integral subscript 0 superscript straight pi over 2 end superscript log open parentheses fraction numerator sin squared straight theta over denominator cos squared straight theta end fraction close parentheses. space sin space 2 straight theta space dθ
equals space integral subscript 0 superscript straight pi over 2 end superscript space log space tan squared straight theta. space space sin space 2 straight theta space dθ space equals space 2 integral subscript 0 superscript straight pi over 2 end superscript log space tanθ. space sin space 2 straight theta space dθ
                  equals 2 integral subscript 0 superscript straight pi over 2 end superscript log space tan open parentheses straight pi over 2 minus straight theta close parentheses space sin space 2 space open parentheses straight pi over 2 minus straight theta close parentheses space dθ space space space space space open square brackets space because space space 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 2 integral subscript 0 superscript straight pi over 2 end superscript space log space cotθ. space sin space 2 straight theta space dθ space equals space 2 integral subscript 0 superscript straight pi over 2 end superscript log open parentheses fraction numerator 1 over denominator tan space straight theta end fraction close parentheses. space sin space 2 straight theta space dθ
space equals space 2 integral subscript 0 superscript straight pi over 2 end superscript left parenthesis log space 1 space minus space log space tanθ right parenthesis space. space sin space 2 straight theta space dθ
rightwards double arrow space space space space straight I space equals space minus 2 space integral subscript 0 superscript straight pi over 2 end superscript log space tanθ comma space sin space 2 straight theta space dθ
therefore space space space space space space straight I space equals space minus 1 comma space space space space space space space space space space rightwards double arrow space space space space 2 space space straight I space equals space space 0 space space space space space space space space space space rightwards double arrow space space space straight I space space equals space 0
                  

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