Question

Show that:
$integral subscript 0 superscript straight pi over 2 end superscript fraction numerator dx over denominator 1 plus square root of tan space straight x end root end fraction space equals space straight pi over 4$

Solution

Let I = $integral subscript 0 superscript straight pi over 2 end superscript fraction numerator dx over denominator 1 plus square root of tan space straight x end root end fraction space equals space integral fraction numerator square root of cos space straight x end root over denominator square root of sin space straight x end root plus square root of cos space straight x end root end fraction dx$ ...(1)
Then I = $integral subscript 0 superscript straight pi over 2 end superscript fraction numerator square root of cos space open parentheses begin display style straight pi over 2 end style minus straight x close parentheses end root over denominator square root of cos space open parentheses begin display style straight pi over 2 end style minus straight x close parentheses end root plus square root of sin space open parentheses begin display style straight pi over 2 end style minus straight x close parentheses end root end fraction dx$
$open square brackets because space space integral subscript 0 superscript straight a straight f left parenthesis straight x right parenthesis space dx space equals space integral subscript 0 superscript straight a straight f left parenthesis straight a minus straight x right parenthesis space dx close square brackets$
$therefore space space space straight I space equals space integral subscript 0 superscript straight pi over 2 end superscript fraction numerator square root of sin space straight x end root over denominator square root of sin space straight x end root plus square root of cosx end fraction dx$ ...(2)
Adding (1) and (2), we get,
$2 space straight I space equals space integral subscript 0 superscript straight pi over 2 end superscript open square brackets fraction numerator square root of cos space straight x end root over denominator square root of cos space straight x space end root plus square root of sin space straight x end root end fraction plus fraction numerator square root of sin space straight x end root over denominator square root of sin space straight x end root plus square root of cos space straight x end root end fraction close square brackets dx space space space space space space equals space integral subscript 0 superscript straight pi over 2 end superscript fraction numerator square root of cos space straight x end root plus square root of sin space straight x end root over denominator square root of cos space straight x end root plus square root of sin space straight x end root end fraction dx space equals space integral subscript 0 superscript straight pi over 2 end superscript 1. space dx space equals space open square brackets straight x close square brackets subscript 0 superscript straight pi over 2 end superscript space equals space straight pi over 2 minus 0$
$therefore space space 2 space straight I space space equals space straight pi over 2 space space space space space space space rightwards double arrow space space space space space space straight I space equals space straight pi over 4$

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Integrals

Show that:
$integral subscript 0 superscript straight pi over 2 end superscript fraction numerator dx over denominator 1 plus square root of tan space straight x end root end fraction space equals space straight pi over 4$

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