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Integrals

Question
CBSEENMA12032536

Evaluate integral subscript 0 superscript straight pi fraction numerator dx over denominator 5 plus 4 space cosx end fraction.

Solution

Let I = integral subscript 0 superscript straight pi fraction numerator dx over denominator 5 plus 4 space cosx end fraction
Put   tan space straight x over 2 space equals space straight t space space space space or space space space space straight x over 2 space equals space tan to the power of negative 1 end exponent straight t space space space space space space or space space straight x space equals space 2 space tan to the power of negative 1 end exponent straight t space space space space space rightwards double arrow space space space space dx space equals space fraction numerator 2 over denominator 1 plus straight t squared end fraction dt
                  cosx space equals space fraction numerator 1 minus tan squared 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 1 minus straight t squared over denominator 1 plus straight t squared end fraction
When x = 0, t = tan 0 = 0
When straight x equals space straight pi over 2 comma space space straight t space equals space tan straight pi over 2 space equals space infinity
therefore      I = 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 5 plus 4 open parentheses begin display style fraction numerator 1 minus straight t squared over denominator 1 plus straight t squared end fraction end style close parentheses end fraction space equals space integral subscript 0 superscript infinity fraction numerator 2 space dt over denominator 5 space left parenthesis 1 plus straight t squared right parenthesis space plus space 4 left parenthesis 1 minus straight t squared right parenthesis end fraction
         equals space 2 integral subscript 0 superscript infinity fraction numerator dt over denominator straight t squared plus 9 end fraction space equals space 2 integral subscript 0 superscript infinity fraction numerator dt over denominator straight t squared plus left parenthesis 3 right parenthesis squared end fraction
equals 2. space 1 third open square brackets tan to the power of negative 1 end exponent straight t over 3 close square brackets subscript 0 superscript infinity space equals space 2 over 3 open square brackets tan to the power of negative 1 end exponent infinity space minus space tan to the power of negative 1 end exponent 0 close square brackets space equals space 2 over 3 open square brackets straight pi over 2 minus 0 close square brackets space equals space straight pi over 3
             

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