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Integrals

Question
CBSEENMA12032580

Evaluate  integral subscript negative 8 end subscript superscript 8 space log space open parentheses fraction numerator 2 plus straight x over denominator 2 minus straight x end fraction close parentheses dx.

Solution
integral subscript negative straight a end subscript superscript straight a straight f left parenthesis straight x right parenthesis space dx space equals space minus integral subscript 0 superscript straight a straight f left parenthesis straight x right parenthesis space dx space plus space integral subscript 0 superscript straight a straight f left parenthesis straight x right parenthesis space dx
rightwards double arrow space space space integral subscript negative straight a end subscript superscript straight a space straight f left parenthesis straight x right parenthesis space dx space equals space 0
Let                         straight I space equals space integral subscript negative 8 end subscript superscript 8 space log space open parentheses fraction numerator 2 plus straight x over denominator 2 minus straight x end fraction close parentheses dx

Let           straight f left parenthesis straight x right parenthesis space equals space log open parentheses fraction numerator 2 plus straight x over denominator 2 minus straight x end fraction close parentheses
therefore space space space straight f left parenthesis negative straight x right parenthesis space equals space log open parentheses fraction numerator 2 minus straight x over denominator 2 plus straight x end fraction close parentheses space equals space log open parentheses fraction numerator 2 plus straight x over denominator 2 minus straight x end fraction close parentheses to the power of negative 1 end exponent space equals space minus log open parentheses fraction numerator 2 plus straight x over denominator 2 minus straight x end fraction close parentheses space equals space minus straight f left parenthesis straight x right parenthesis
therefore space space space space space space straight f left parenthesis straight x right parenthesis  is an odd function of x.
therefore space space space space space straight I equals space 0                           open square brackets because space integral subscript straight a superscript straight a straight f left parenthesis straight x right parenthesis space dx space equals space 0 space if space straight f left parenthesis straight x right parenthesis space is space an space odd space function space of space straight x close square brackets

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