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Integrals

Question
CBSEENMA12032539

Evaluate the following:
integral subscript 0 superscript straight pi fraction numerator 1 over denominator 6 minus cosx end fraction dx

Solution

Let I = integral subscript 0 superscript straight pi fraction numerator 1 over denominator 6 minus cosx end fraction dx
Put tan space straight x over 2 space equals space straight t comma space space space or space space straight x over 2 space equals space tan to the power of negative 1 end exponent straight t comma space space space or space space straight x space equals space 2 space tan to the power of negative 1 end exponent space straight t space space rightwards double arrow space space dx space equals space fraction numerator 2 over denominator 1 plus straight t squared end fraction dt
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
Also  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 fraction numerator 1 minus straight t squared over denominator 1 plus straight t squared end fraction
 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 6 minus begin display style fraction numerator 1 minus straight t squared over denominator 1 plus straight t squared end fraction end style end fraction space equals integral subscript 0 superscript infinity fraction numerator 2 over denominator 6 plus 6 straight t squared minus 1 plus straight t squared end fraction dt space equals space 2 integral subscript 0 superscript infinity fraction numerator 1 over denominator 7 straight t squared plus 5 end fraction dt
                  equals space 2 over 7 integral subscript 0 superscript infinity fraction numerator 1 over denominator straight t squared plus open parentheses begin display style fraction numerator square root of 5 over denominator 7 end fraction end style close parentheses squared end fraction dt space equals space 2 over 7. fraction numerator 1 over denominator square root of begin display style 5 over 7 end style end root end fraction open square brackets tan to the power of negative 1 end exponent open parentheses fraction numerator straight t over denominator square root of begin display style 5 over 7 end style end root end fraction close parentheses close square brackets subscript 0 superscript infinity
equals space fraction numerator 2 over denominator square root of 35 end fraction left square bracket tan to the power of negative 1 end exponent infinity space minus space tan to the power of negative 1 end exponent 0 right square bracket space equals space fraction numerator 2 over denominator square root of 35 end fraction open square brackets straight pi over 2 minus 0 close square brackets space equals space fraction numerator straight pi over denominator square root of 35 end fraction

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