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By using the properties of definite integrals, evaluate the following:
$integral subscript 0 superscript straight pi fraction numerator straight x space dx over denominator straight a squared space cos squared straight x plus straight b squared space sin squared straight x end fraction$

Let I = $integral subscript 0 superscript straight pi fraction numerator straight x space dx over denominator straight a squared cos squared straight x plus straight b squared sin squared straight x end fraction space space$ ........(1)

$therefore space space space space space thin space straight I space equals space integral subscript 0 superscript straight pi fraction numerator left parenthesis straight pi minus straight x right parenthesis dx over denominator straight a squared cos squared left parenthesis straight pi minus straight x right parenthesis plus straight b squared sin squared left parenthesis straight pi minus straight x right parenthesis end fraction space equals space integral subscript 0 superscript straight pi fraction numerator left parenthesis straight pi minus straight x right parenthesis space dx over denominator straight a squared cos squared straight x plus straight b squared sin squared straight x end fraction dx$
$equals straight pi integral subscript 0 superscript straight pi fraction numerator 1 over denominator straight a squared cos squared straight x plus straight b squared sin squared straight x end fraction dx space minus integral subscript 0 superscript straight pi fraction numerator xdx over denominator straight a squared space cos squared straight x plus straight b squared space sin squared straight x end fraction$

$therefore space space space space space straight I space equals space straight pi integral subscript 0 superscript straight pi fraction numerator 1 over denominator straight a squared cos squared straight x plus straight b squared sin squared straight x end fraction dx minus 1$ $open square brackets because space of space left parenthesis 1 right parenthesis close square brackets$

$therefore space space space 2 space straight I space equals space 2 integral subscript 0 superscript straight pi over 2 end superscript fraction numerator 1 over denominator straight a squared space cos squared straight x plus straight b squared space sin squared straight x end fraction dx$

$therefore space space 2 space space straight I space space equals space 2 integral subscript 0 superscript straight pi fraction numerator begin display style fraction numerator 1 over denominator cos squared straight x end fraction end style dx over denominator straight a squared plus straight b squared begin display style fraction numerator sin squared straight x over denominator cos squared straight x end fraction end style end fraction$
$therefore space space space space straight I space equals space straight pi integral subscript 0 superscript straight pi over 2 end superscript fraction numerator sec squared straight x space dx over denominator straight a squared plus straight b squared space tan squared straight x end fraction$

Put tan x = t, ∴ sec² x dx = dt When x = 0, t = tan 0 = 0
When $straight x space equals space straight pi over 2 comma space space space straight t space equals space tan straight pi over 2 space equals space infinity$
$therefore space space space space space straight I space equals space straight pi integral subscript 0 superscript infinity fraction numerator dt over denominator straight a squared plus straight b squared straight t squared end fraction space equals space straight pi over straight b squared integral subscript 0 superscript infinity fraction numerator 1 over denominator open parentheses begin display style straight a over straight b end style close parentheses squared plus straight t squared end fraction dt$
$equals space straight pi over straight b squared. fraction numerator 1 over denominator begin display style straight a over straight b end style end fraction open square brackets tan to the power of negative 1 end exponent open parentheses fraction numerator straight t over denominator begin display style straight a over straight b end style end fraction close parentheses close square brackets subscript 0 superscript infinity space equals space fraction numerator straight pi over denominator straight a space straight b end fraction left parenthesis tan to the power of negative 1 end exponent infinity space minus space tan to the power of negative 1 end exponent 0 right parenthesis space equals space straight pi over ab open parentheses straight pi over 2 minus 0 close parentheses$
$therefore space space space space straight I space equals space fraction numerator straight pi squared over denominator 2 ab end fraction$

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