Question

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

Solution

Let I = $integral subscript 0 superscript straight a fraction numerator 1 over denominator straight x plus square root of straight a squared minus straight x squared end root end fraction$
Put x = a sin θ,    ∴ dx = a cos θ dθ
When x = 0, a sin θ = 0 ⇒ sin θ = 0 ⇒ θ = 0
when x = a, $straight a space sin space straight theta space equals space straight a space space space space space space space space space space rightwards double arrow space space space space sin space straight theta space equals space 1 space space space space rightwards double arrow space space space straight theta space equals space straight pi over 2$
$therefore space space space space straight I space equals space integral subscript 0 superscript straight pi over 2 end superscript fraction numerator 1 over denominator straight a space sinθ space plus space square root of straight a squared minus straight a squared space sin squared space straight theta end root end fraction. space straight a space cosθ space dθ$
$equals space integral subscript 0 superscript straight pi over 2 end superscript fraction numerator cos space straight theta space dθ over denominator sin space straight theta space plus space cos space straight theta end fraction space equals space 1 half integral subscript 0 superscript straight pi over 2 end superscript fraction numerator 2 space cos space straight theta over denominator sin space straight theta space plus space cos space straight theta end fraction dθ$
   $equals space 1 half integral subscript 0 superscript straight pi over 2 end superscript fraction numerator left parenthesis cosθ space plus space sinθ right parenthesis space plus space left parenthesis cosθ space minus space sinθ right parenthesis over denominator cosθ plus sinθ end fraction dθ space equals 1 half integral subscript 0 superscript straight pi over 2 end superscript open parentheses 1 plus fraction numerator cosθ minus sinθ over denominator cosθ plus sinθ end fraction close parentheses dθ$

   $equals space 1 half open square brackets straight theta plus log open vertical bar cos space straight theta space plus space sin space straight theta close vertical bar close square brackets subscript 0 superscript straight pi over 2 end superscript equals space 1 half open square brackets open parentheses straight pi over 2 plus log space open vertical bar cos straight pi over 2 plus sin straight pi over 2 close vertical bar close parentheses space minus space left parenthesis 0 plus log open vertical bar cos space 0 plus space sin space 0 close vertical bar right parenthesis close square brackets equals space 1 half open square brackets open parentheses straight pi over 2 plus log space open vertical bar 0 plus 1 close vertical bar close parentheses space minus space left parenthesis 0 plus log space open vertical bar 1 plus 0 close vertical bar right parenthesis close square brackets space equals space 1 half open square brackets straight pi over 2 close square brackets space equals straight pi over 4$

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Integrals

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

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