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 Multiple Choice QuestionsShort Answer Type

201.

Evaluate:
integral subscript 0 superscript straight pi fraction numerator straight x space over denominator 1 plus sinx end fraction dx space equals space straight pi

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202.

Show that:
integral subscript 0 superscript straight pi over 2 end superscript fraction numerator straight x space dx over denominator sinx space plus space cosx end fraction space equals space fraction numerator straight pi over denominator 2 square root of 2 end fraction log left parenthesis square root of 2 plus 1 right parenthesis

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 Multiple Choice QuestionsLong Answer Type

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203.

Show that:
integral subscript 0 superscript straight pi over 2 end superscript fraction numerator sin squared straight x over denominator sin space straight x space plus space cos space straight x end fraction space equals space fraction numerator 1 over denominator square root of 2 end fraction log left parenthesis square root of 2 plus 1 right parenthesis


Let I = integral subscript 0 superscript straight pi over 2 end superscript fraction numerator sin squared straight x over denominator sinx plus cosx end fraction dx                          ...(1)
therefore space space space space space space space space straight I space equals space integral subscript 0 superscript straight pi over 2 end superscript space fraction numerator sin squared open parentheses begin display style straight pi over 2 end style minus straight x close parentheses over denominator sin open parentheses begin display style straight pi over 2 end style minus straight x close parentheses plus cos open parentheses begin display style straight pi over 2 end style minus straight x close parentheses end fraction dx space equals space integral subscript 0 superscript straight pi over 2 end superscript fraction numerator cos squared straight x over denominator cosx plus sinx end fraction dx  ...(2)
Adding (1) and (2), we get,
            2 space straight I space equals space integral subscript 0 superscript straight pi over 2 end superscript open square brackets fraction numerator sin squared straight x over denominator sinx plus cosx end fraction plus fraction numerator cos squared straight x over denominator sinx plus cosx end fraction close square brackets dx space equals space integral subscript 0 superscript straight pi over 2 end superscript fraction numerator sin squared plus cos squared straight x over denominator sinx plus cosx end fraction dx
therefore space space space space space space space space space space space space space space 2 space straight I space equals space integral subscript 0 superscript straight pi over 2 end superscript fraction numerator 1 over denominator sinx plus cosx end fraction dx space space space space space space space space space space space space space space space space space space space space... left parenthesis 3 right parenthesis
Let straight I subscript 1 space equals space integral subscript 0 superscript straight pi over 2 end superscript fraction numerator 1 over denominator cosx plus sinx end fraction dx
Put tan straight x over 2 space equals space straight t space space space space or space space straight x over 2 space equals space tan to the power of negative 1 end exponent straight t 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 space rightwards double arrow space space dx 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 space space and space 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 space equals space straight pi over 2 comma space space space straight t space equals space tan straight pi over 4 space equals space 1
therefore space space space straight I subscript 1 space equals space integral subscript 0 superscript 1 fraction numerator begin display style fraction numerator 1 over denominator 1 plus straight t squared end fraction end style dt over denominator begin display style fraction numerator 1 minus straight t squared over denominator 1 plus straight t squared end fraction end style 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 2 integral subscript 0 superscript 1 fraction numerator dt over denominator 1 minus straight t squared plus 2 straight t end fraction space equals space 2 integral subscript 0 superscript 1 fraction numerator dt over denominator 2 minus left parenthesis straight t squared minus 2 straight t right parenthesis end fraction space equals 2 integral subscript 0 superscript 1 fraction numerator dt over denominator 2 minus left parenthesis straight t squared minus 2 straight t plus 1 right parenthesis end fraction
           
therefore space space space space space integral subscript 0 superscript straight pi over 2 end superscript fraction numerator dx over denominator cosx plus sinx end fraction space equals space fraction numerator 2 over denominator square root of 2 end fraction log left parenthesis square root of 2 plus 1 right parenthesis
therefore space space space space from space left parenthesis 3 right parenthesis comma space space 2 space straight I space equals space fraction numerator 2 over denominator square root of 2 end fraction log left parenthesis square root of 2 plus 1 right parenthesis
therefore space space straight I space equals space fraction numerator 1 over denominator square root of 2 end fraction log left parenthesis square root of 2 plus 1 right parenthesis

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 Multiple Choice QuestionsShort Answer Type

204.

Show that:
integral subscript 0 superscript straight pi fraction numerator straight x space tanx over denominator secx space cosecx end fraction space equals space straight pi squared over 4


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205.

Show that:
integral subscript 0 superscript straight pi fraction numerator straight x space tanx over denominator secx plus tanx end fraction dx space equals space straight pi open parentheses straight pi over 2 minus 1 close parentheses

119 Views

 Multiple Choice QuestionsLong Answer Type

206.

Show that:
integral subscript 0 superscript 1 log space open parentheses 1 over straight x minus 1 close parentheses dx space equals space 0

139 Views

 Multiple Choice QuestionsShort Answer Type

207.

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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208.

Show that:
integral subscript 0 superscript straight pi space straight x space. space log space sinx space dx space equals space minus straight pi squared over 2 log space 2

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209.

Show that:
integral subscript 0 superscript straight pi over 2 end superscript open parentheses fraction numerator straight theta over denominator sin space straight theta end fraction close parentheses squared space dθ space equals space straight pi space log space 2

153 Views

210.

Show that:
integral subscript 0 superscript straight pi fraction numerator straight x space sinx over denominator 1 plus cos squared straight x end fraction dx space equals space straight pi squared over 4


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