Question

By using the properties of definite integrals, evaluate the following integral:
$integral subscript 0 superscript straight pi fraction numerator dx over denominator 1 plus sinx end fraction space equals space 2$

Solution

Let I = $integral subscript 0 superscript straight pi fraction numerator dx over denominator 1 plus sinx end fraction space equals space integral subscript 0 superscript straight pi fraction numerator dx over denominator 1 plus sinx end fraction cross times fraction numerator 1 minus sinx over denominator 1 minus sinx end fraction dx space equals space straight x integral subscript 0 superscript straight pi fraction numerator 1 minus sinx over denominator 1 minus sin squared straight x end fraction dx$
$equals space integral subscript 0 superscript straight pi fraction numerator 1 minus sinx over denominator cos squared straight x end fraction dx space equals space integral subscript 0 superscript straight pi open parentheses fraction numerator 1 over denominator cos squared straight x end fraction minus fraction numerator sinx over denominator cos squared straight x end fraction close parentheses dx space equals space integral subscript 0 superscript straight pi open parentheses fraction numerator 1 over denominator cos squared straight x end fraction minus 1 over cosx. sinx over cosx close parentheses dx$
$space equals space integral subscript 0 superscript straight pi left parenthesis sec squared straight x minus secx space tanx right parenthesis space dx space equals space open square brackets tanx space minus secx close square brackets subscript 0 superscript straight pi space equals space open square brackets left parenthesis tanπ minus secπ right parenthesis space minus space left parenthesis tan space 0 space minus space sec space 0 right parenthesis close square brackets space equals space open square brackets left parenthesis 0 plus 1 right parenthesis space minus space left square bracket 0 minus 1 right parenthesis close square brackets space equals space 2$

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Integrals

By using the properties of definite integrals, evaluate the following integral:
$integral subscript 0 superscript straight pi fraction numerator dx over denominator 1 plus sinx end fraction space equals space 2$

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