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Integrals

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

Let I = $integral subscript 0 superscript straight pi over 2 end superscript fraction numerator xdx over denominator sinx plus cosx end fraction$ ...(1)
$therefore space space space space straight I space equals space integral subscript 0 superscript straight pi over 2 end superscript fraction numerator begin display style straight pi over 2 end style minus straight x over denominator sin open parentheses begin display style straight pi over 2 end style minus straight x close parentheses plus cos open parentheses begin display style straight pi over 2 end style minus straight x close parentheses end fraction dx space space space space space space open square brackets because space integral subscript 0 superscript straight a straight f left parenthesis straight x right parenthesis space dx space space equals space integral subscript 0 superscript straight a straight f left parenthesis straight a minus straight x right parenthesis space dx close square brackets$

$equals space integral subscript 0 superscript straight pi over 2 end superscript fraction numerator begin display style straight pi over 2 minus straight x end style over denominator cosx plus sinx end fraction dx space equals space straight pi over 2 integral subscript 0 superscript straight pi over 2 end superscript fraction numerator dx over denominator cosx plus sinx end fraction space minus space integral subscript 0 superscript straight pi over 2 end superscript fraction numerator xdx over denominator cosx plus sinx end fraction$

$therefore space space space straight I space equals space space straight pi over 2 integral subscript 0 superscript straight pi over 2 end superscript fraction numerator dx over denominator cosx plus sinx end fraction space minus 1$ [ $because space space of space left parenthesis 1 right parenthesis$ ]

$therefore space space space space 2 space straight I space equals space straight pi over 2 integral subscript 0 superscript straight pi over 2 end superscript fraction numerator 1 over denominator cosx plus sinx end fraction dx$ ...(2)

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 $tan straight x over 2 space equals space straight t space space or space space straight x over 2 space equals space tan to the power of 1 straight t space space or space space straight x space equals space 2 space tan to the power of 1 straight t space space space space rightwards double arrow space space space space space dx space space equals fraction numerator 2 over denominator 1 plus straight t squared end fraction dt$

$sinx space equals space fraction numerator 2 space tan begin display style straight x over 2 end style over denominator 1 plus tan squared begin display style straight x over 2 end style end fraction space equals fraction numerator 2 space straight t over denominator 1 plus straight t squared end fraction space and space cosx space equals space fraction numerator 1 minus tan squared begin display style straight x over 2 end style over denominator 1 plus tan squared begin display style straight x over 2 end style end fraction space equals fraction numerator 1 minus straight t squared over denominator 1 plus straight t squared end fraction space$

When x = 0, t = tan 0 = 0

When $straight x space equals space straight pi over 2 comma space space straight t space equals space tan straight pi over 4 space equals space 1$
$therefore space space space space space space straight I subscript 1 space equals space integral subscript 0 superscript 1 fraction numerator begin display style fraction numerator 1 over denominator 1 plus straight t squared end fraction end style dt over denominator begin display style fraction numerator 1 minus straight t squared over denominator 1 plus straight t squared end fraction end style plus begin display style fraction numerator 2 space straight t over denominator 1 plus straight t squared end fraction end style end fraction space equals space 2 integral subscript 0 superscript 1 fraction numerator dt over denominator 1 minus straight t squared plus 2 straight t end fraction space equals space 2 integral subscript 0 superscript 1 fraction numerator dt over denominator 2 minus left parenthesis straight t squared minus 2 straight t plus 1 right parenthesis end fraction$
$equals space 2 integral subscript 0 superscript 1 fraction numerator dt over denominator left parenthesis square root of 2 right parenthesis squared minus left parenthesis straight t minus 1 right parenthesis squared end fraction space equals 2 cross times fraction numerator 1 over denominator 2 square root of 2 end fraction open square brackets log space open vertical bar fraction numerator square root of 2 plus left parenthesis straight t minus 1 right parenthesis over denominator square root of 2 minus left parenthesis straight t minus 1 right parenthesis end fraction close vertical bar close square brackets subscript 0 superscript 1 equals space fraction numerator 1 over denominator square root of 2 end fraction open square brackets log space fraction numerator square root of 2 over denominator square root of 2 end fraction minus log fraction numerator square root of 2 minus 1 over denominator square root of 2 plus 1 end fraction close square brackets space equals space minus fraction numerator 1 over denominator square root of 2 end fraction log fraction numerator square root of 2 minus 1 over denominator square root of 2 plus 1 end fraction equals space fraction numerator 1 over denominator square root of 2 end fraction log fraction numerator square root of 2 plus 1 over denominator square root of 2 minus 1 end fraction equals fraction numerator 1 over denominator square root of 2 end fraction log open parentheses fraction numerator square root of 2 plus 1 over denominator square root of 2 minus 1 end fraction cross times fraction numerator square root of 2 plus 1 over denominator square root of 2 plus 1 end fraction close parentheses equals space fraction numerator 1 over denominator square root of 2 end fraction log open square brackets fraction numerator left parenthesis square root of 2 plus 1 right parenthesis squared over denominator 2 minus 1 end fraction close square brackets 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 squared 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$
$therefore space space space space space integral subscript 0 superscript straight pi over 2 end superscript fraction numerator dx over denominator cosx plus sinx end fraction space equals space fraction numerator 2 over denominator square root of 2 end fraction log left parenthesis square root of 2 plus 1 right parenthesis therefore space space space space space space space equals space fraction numerator straight pi over denominator 2 square root of 2 end fraction log left parenthesis square root of 2 plus 1 right parenthesis$

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Long Answer Type

203.

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

146 Views

Short Answer Type

204.

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

101 Views

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

Long 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

Short Answer Type

207.

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

135 Views

208.

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

100 Views

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

125 Views

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