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Integrals

Question
CBSEENMA12032540

Evaluate the following:
integral subscript 0 superscript straight pi fraction numerator dx over denominator 1 plus sin space straight x end fraction


Solution

Let I = integral subscript 0 superscript straight pi fraction numerator dx over denominator 1 plus sinx end fraction
Put tan straight x over 2 space equals space straight t comma space space space or space space straight x over 2 equals space tan to the power of negative 1 end exponent straight t comma space space or space space straight x space equals space 2 tan to the power of negative 1 end exponent straight t space space rightwards double arrow space space space dx space space equals fraction numerator 2 over denominator 1 plus straight t squared end fraction dt
     sinx space equals space fraction numerator 2 tan begin display style straight x over 2 end style over denominator 1 plus tan squared begin display style straight x over 2 end style end fraction space equals space fraction numerator 2 straight t over denominator 1 plus straight t squared end fraction
When x = 0. t = tan 0 = 0
When straight x space equals space straight pi comma space space space straight t space equals space tan straight pi over 2 space rightwards arrow space infinity
therefore              straight I space equals space integral subscript 0 superscript infinity fraction numerator begin display style fraction numerator 2 over denominator 1 plus straight t squared end fraction end style dt over denominator 1 plus begin display style fraction numerator 2 straight t over denominator 1 plus straight t squared end fraction end style end fraction space equals space integral subscript 0 superscript infinity fraction numerator 2 over denominator 1 plus straight t squared plus 2 straight t end fraction dt space equals space 2 integral subscript 0 superscript infinity fraction numerator dt over denominator left parenthesis straight t plus 1 right parenthesis squared end fraction space equals 2 integral subscript 0 superscript infinity left parenthesis straight t plus 1 right parenthesis to the power of negative 2 end exponent dt
                     equals space 2 open square brackets fraction numerator left parenthesis straight t plus 1 right parenthesis to the power of negative 1 end exponent over denominator negative 1 end fraction close square brackets subscript 0 superscript infinity space equals space minus 2 open square brackets fraction numerator 1 over denominator straight t plus 1 end fraction close square brackets subscript 0 superscript infinity space equals space minus 2 open square brackets Lt with straight t rightwards arrow infinity below fraction numerator 1 over denominator straight t plus 1 end fraction minus fraction numerator 1 over denominator 0 plus 1 end fraction close square brackets space equals space minus 2 left square bracket 0 minus 1 right square bracket space equals space 2

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