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Integrals

Short Answer Type

201.

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

119 Views

202.

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

120 Views

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

Let I = $integral subscript 0 superscript 1 log space open parentheses 1 over straight x minus 1 close parentheses dx$
Put x = cos² θ, θ dx = 2 cos θ (– sin θ) dθ = – 2 sin θ cos θ dθ = – sin 2θ dθ

When x = 0, $cos squared straight theta space equals space 0 space rightwards double arrow space space space straight theta space equals space straight pi over 2$
When x = 1, cos² θ = 1 ⇒ θ = 0
$therefore space space space straight I space equals space minus integral subscript straight pi over 2 end subscript superscript 0 log open parentheses fraction numerator 1 over denominator cos squared straight theta end fraction minus 1 close parentheses space sin space 2 straight theta space dθ space equals space integral subscript 0 superscript straight pi over 2 end superscript log space open parentheses fraction numerator 1 minus cos squared straight theta over denominator cos squared straight theta end fraction close parentheses space sin space 2 straight theta space dθ$
   $open square brackets because space space space integral subscript straight a superscript straight b straight f left parenthesis straight x right parenthesis space dx space equals space minus integral subscript straight b superscript straight a straight f left parenthesis straight x right parenthesis space dx close square brackets$
   $equals space integral subscript 0 superscript straight pi over 2 end superscript log open parentheses fraction numerator sin squared straight theta over denominator cos squared straight theta end fraction close parentheses. space sin space 2 straight theta space dθ equals space integral subscript 0 superscript straight pi over 2 end superscript space log space tan squared straight theta. space space sin space 2 straight theta space dθ space equals space 2 integral subscript 0 superscript straight pi over 2 end superscript log space tanθ. space sin space 2 straight theta space dθ$

   $equals 2 integral subscript 0 superscript straight pi over 2 end superscript log space tan open parentheses straight pi over 2 minus straight theta close parentheses space sin space 2 space open parentheses straight pi over 2 minus straight theta close parentheses space dθ space space space space space open square brackets space because space space space integral subscript 0 superscript straight a straight f left parenthesis straight x right parenthesis space dx space equals space integral subscript 0 superscript straight a straight f left parenthesis straight a minus straight x right parenthesis space dx close square brackets$

   $equals 2 integral subscript 0 superscript straight pi over 2 end superscript space log space cotθ. space sin space 2 straight theta space dθ space equals space 2 integral subscript 0 superscript straight pi over 2 end superscript log open parentheses fraction numerator 1 over denominator tan space straight theta end fraction close parentheses. space sin space 2 straight theta space dθ space equals space 2 integral subscript 0 superscript straight pi over 2 end superscript left parenthesis log space 1 space minus space log space tanθ right parenthesis space. space sin space 2 straight theta space dθ$

$rightwards double arrow space space space space straight I space equals space minus 2 space integral subscript 0 superscript straight pi over 2 end superscript log space tanθ comma space sin space 2 straight theta space dθ$
$therefore space space space space space space straight I space equals space minus 1 comma space space space space space space space space space space rightwards double arrow space space space space 2 space space straight I space equals space space 0 space space space space space space space space space space rightwards double arrow space space space straight I space space equals space 0$

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