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Integrals

Question
CBSEENMA12032498

Prove that: integral subscript negative straight a end subscript superscript straight a square root of fraction numerator straight a minus straight x over denominator straight a plus straight x end fraction end root space dx space equals space straight a space straight pi.

Solution

Let I = integral subscript negative straight a end subscript superscript straight a space square root of fraction numerator straight a minus straight x over denominator straight a plus straight x end fraction end root dx
Put x = a cos θ , ∴ dx = –a sin θ dθ
When x = a, a = a cos θ ⇒ 1 = cos θ    ⇒ θ = 0
When x = –a, –a = a cos θ ⇒ – 1 = cos θ ⇒ θ = straight pi
therefore                               straight I space equals space integral subscript straight pi superscript 0 square root of fraction numerator straight a minus acosθ over denominator straight a plus acosθ end fraction end root space space space space left parenthesis negative straight a space sin space straight theta space dθ right parenthesis
                         equals space straight a space integral subscript 0 superscript straight pi square root of fraction numerator 1 minus cosθ over denominator 1 plus cosθ end fraction end root. space space sin space straight theta space dθ space equals space straight a space integral subscript 0 superscript straight pi square root of fraction numerator 2 sin squared begin display style straight theta over 2 end style over denominator 2 cos squared begin display style straight theta over 2 end style end fraction end root. space sin space straight theta space dθ
    equals space straight a space integral subscript 0 superscript straight pi fraction numerator sin space begin display style straight theta over 2 end style over denominator cos begin display style straight theta over 2 end style end fraction. space 2 space sin space straight theta over 2 space cos straight theta over 2 space dθ space equals space straight a space integral subscript 0 superscript straight pi space 2 space sin squared straight theta over 2 space dθ
   equals space straight a integral subscript 0 superscript straight pi left parenthesis 1 minus cos space straight theta right parenthesis dθ space equals space straight a space open square brackets straight theta minus sinθ close square brackets subscript 0 superscript straight pi
   equals space straight a open square brackets left parenthesis straight pi minus sin space straight pi right parenthesis space minus space left parenthesis 0 minus sin space 0 right parenthesis close square brackets
equals space straight a open square brackets left parenthesis straight pi minus 0 right parenthesis space minus space left parenthesis 0 minus 0 right parenthesis close square brackets space equals aπ.

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