Question

Evaluate $integral subscript straight pi over 6 end subscript superscript straight pi over 3 end superscript fraction numerator dx over denominator 1 plus square root of tanx end fraction$

Solution

Let I = $integral subscript straight pi over 6 end subscript superscript straight pi over 3 end superscript fraction numerator 1 over denominator 1 plus square root of tan space straight x end root end fraction dx space equals space integral subscript straight pi over 6 end subscript superscript straight pi over 3 end superscript fraction numerator 1 over denominator 1 plus begin display style fraction numerator square root of sin space straight x end root over denominator square root of cosx end fraction end style end fraction dx$
$therefore space space space straight I space equals space integral subscript straight pi over 6 end subscript superscript straight pi over 3 end superscript fraction numerator square root of cos space straight x end root over denominator square root of cos space straight x end root plus square root of sin space straight x end root end fraction dx$ ....(1)
Then I = $integral subscript straight pi over 6 end subscript superscript straight pi over 3 end superscript fraction numerator square root of cos space open parentheses begin display style straight pi over 3 end style plus begin display style straight pi over 6 end style minus straight x close parentheses end root over denominator square root of cos space open parentheses begin display style straight pi over 3 end style plus begin display style straight pi over 6 end style minus straight x close parentheses end root plus square root of sin space open parentheses begin display style straight pi over 3 end style plus begin display style straight pi over 6 end style minus straight x close parentheses end root end fraction dx$
$open square brackets because space space integral subscript straight a superscript straight b straight f left parenthesis straight x right parenthesis space dx space equals space integral subscript straight a superscript straight b straight f left parenthesis straight a plus straight b minus straight x right parenthesis space dx close square brackets$
$equals space integral subscript straight pi over 6 end subscript superscript straight pi over 3 end superscript fraction numerator square root of cos space open parentheses begin display style straight pi over 2 end style minus straight x close parentheses end root over denominator square root of cos space open parentheses begin display style straight pi over 2 end style minus straight x close parentheses end root plus square root of sin space open parentheses begin display style straight pi over 2 end style minus straight x close parentheses end root end fraction dx$
$therefore space space space straight I space equals space integral subscript straight pi over 6 end subscript superscript straight pi over 3 end superscript fraction numerator square root of sin space straight x end root over denominator square root of sin space straight x end root plus square root of cos space straight x end root end fraction dx$ ...(2)
Adding (1) and (2), we get.
$2 space straight I space equals space integral subscript straight pi over 6 end subscript superscript straight pi over 3 end superscript open square brackets fraction numerator square root of cos space straight x end root over denominator square root of cos space straight x end root plus square root of sin space straight x end root end fraction plus fraction numerator square root of sin space straight x end root over denominator square root of cosx plus square root of sin space straight x end root end fraction close square brackets dx$
$equals space integral subscript straight pi over 6 end subscript superscript straight pi over 3 end superscript fraction numerator square root of cos space straight x end root plus square root of sin space straight x end root over denominator square root of cos space straight x end root plus square root of sin space straight x end root end fraction dx space equals space integral subscript straight pi over 6 end subscript superscript straight pi over 3 end superscript 1 space dx space equals space open square brackets straight x close square brackets subscript straight pi divided by 6 end subscript superscript straight pi divided by 3 end superscript space equals space straight pi over 3 minus straight pi over 6 space equals space fraction numerator 2 straight pi minus straight pi over denominator 6 end fraction$
$therefore space space 2 space space straight I space equals space straight pi over 6 space space space space rightwards double arrow space space space straight I space space equals space straight pi over 12$

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Integrals

Evaluate $integral subscript straight pi over 6 end subscript superscript straight pi over 3 end superscript fraction numerator dx over denominator 1 plus square root of tanx end fraction$

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