Prove the following:
$integral subscript 0 superscript 1 square root of fraction numerator 1 minus straight x over denominator 1 plus straight x end fraction dx end root space equals space straight pi over 2 minus 1.$

Solution

Let I = $integral subscript 0 superscript 1 square root of fraction numerator 1 minus straight x over denominator 1 plus straight x end fraction end root dx space equals space integral subscript 0 superscript 1 square root of fraction numerator 1 minus straight x over denominator 1 plus straight x end fraction cross times fraction numerator 1 minus straight x over denominator 1 minus straight x end fraction end root dx$
$equals space integral subscript 0 superscript 1 fraction numerator 1 minus straight x over denominator square root of 1 minus straight x squared end root end fraction dx space equals space integral subscript 0 superscript 1 fraction numerator 1 over denominator square root of 1 minus straight x squared end root end fraction dx space minus space integral subscript 0 superscript 1 fraction numerator straight x over denominator square root of 1 minus straight x squared end root end fraction dx equals space integral subscript 0 superscript 1 fraction numerator 1 over denominator square root of 1 minus straight x squared end root end fraction dx plus 1 half integral subscript 0 superscript 1 left parenthesis 1 minus straight x squared right parenthesis space to the power of 1 half end exponent left parenthesis negative 2 straight x right parenthesis space dx$
$equals space open square brackets sin to the power of negative 1 end exponent straight x close square brackets subscript 0 superscript 1 space plus space 1 half open square brackets fraction numerator left parenthesis 1 minus straight x squared right parenthesis to the power of begin display style 1 half end style end exponent over denominator begin display style 1 half end style end fraction close square brackets subscript 0 superscript 1 space equals space open square brackets sin to the power of negative 1 end exponent straight x close square brackets subscript 0 superscript 1 plus open square brackets square root of 1 minus straight x squared end root close square brackets subscript 0 superscript 1$
= $left parenthesis sin to the power of negative 1 end exponent 1 minus sin to the power of negative 1 end exponent 0 right parenthesis space plus space left parenthesis square root of 1 minus 1 end root space minus space square root of 1 minus 0 right parenthesis end root space equals space straight pi over 2 minus 0 plus 0 minus 1 space equals space straight pi over 2 minus 1.$

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Integrals

Prove the following:
$integral subscript 0 superscript 1 square root of fraction numerator 1 minus straight x over denominator 1 plus straight x end fraction dx end root space equals space straight pi over 2 minus 1.$

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