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Integrals

Short Answer Type

191.

Show that:
$integral subscript straight pi over 4 end subscript superscript straight pi over 4 end superscript open vertical bar sin space straight x close vertical bar dx space equals space 2 minus square root of 2$

120 Views

Long Answer Type

192.

Show that:
$integral subscript negative 1 end subscript superscript 3 over 2 end superscript open vertical bar straight x space sinπ space straight x close vertical bar space dx space equals space 3 over straight pi plus 1 over straight pi squared$

152 Views

Short Answer Type

193.

Show that:
$integral subscript 0 superscript 2 straight x square root of 2 minus straight x end root space equals fraction numerator 16 square root of 2 over denominator 15 end fraction$

147 Views

194.

By using the properties of definite integrals, evaluate the following integral:
$integral subscript 0 superscript 1 space straight x space left parenthesis 1 minus straight x right parenthesis to the power of straight n space dx$

108 Views

195.

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

134 Views

196.

Show that:
$integral subscript 0 superscript straight pi over 2 end superscript 2 straight x space log space tanx space dx space equals space 0$

110 Views

197.

By using the properties of definite integrals, evaluate the following integral:
$integral subscript 0 superscript straight pi over 2 end superscript left parenthesis 2 space log space sinx space minus space log space sin space 2 straight x right parenthesis space dx$

410 Views

198.

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

100 Views

199.

Show that:
$integral subscript 0 superscript infinity log space open parentheses straight x plus 1 over straight x close parentheses. space fraction numerator dx over denominator 1 plus straight x squared end fraction space equals space straight pi space log space 2$

132 Views

200.
Show that:
$integral subscript 0 superscript 1 fraction numerator log space left parenthesis 1 plus straight x right parenthesis over denominator 1 plus straight x squared end fraction space equals space straight pi over 8 space log space 2$

Let I = $integral subscript 0 superscript 1 fraction numerator log left parenthesis 1 plus straight x right parenthesis over denominator 1 plus straight x squared end fraction dx$
Put x = tan θ, ∴ dx = sec² θ dθ
When x = 0, tan θ = 0 ⇒ θ = 0
When x = 1, $tan space straight theta space equals space 1 space space space space space space space space rightwards double arrow space space space space straight theta space equals space straight pi over 4$

$therefore space space space straight I space equals space integral subscript 0 superscript straight pi over 4 end superscript space log left parenthesis 1 plus tanθ right parenthesis space dθ space$ ...(1)
$equals space integral subscript 0 superscript straight pi over 4 end superscript log space open square brackets 1 plus tan open parentheses straight pi over 4 minus straight theta close parentheses close square brackets space dθ$ $open square brackets because space integral subscript 0 superscript straight a straight f left parenthesis straight x right parenthesis space dx space equals space integral subscript 0 superscript straight a straight f left parenthesis straight a minus straight x right parenthesis space dx close square brackets$
$equals space integral subscript 0 superscript straight pi over 4 end superscript log open square brackets 1 plus fraction numerator 1 minus tanθ over denominator 1 plus tanθ end fraction close square brackets dθ space equals space integral subscript 0 superscript straight pi over 4 end superscript log open square brackets fraction numerator 1 plus tanθ plus 1 minus tanθ over denominator 1 plus tanθ end fraction close square brackets dθ space equals space integral subscript 0 superscript straight pi over 4 end superscript log open parentheses fraction numerator 2 over denominator 1 plus tanθ end fraction close parentheses dθ$

$equals space integral subscript 0 superscript straight pi over 4 end superscript log space 2 space dθ space space minus integral subscript 0 superscript straight pi over 4 end superscript log left parenthesis 1 plus tanθ right parenthesis space dθ$
$therefore space space space space straight I space equals space log space 2 integral subscript 0 superscript straight pi over 4 end superscript space 1. space dθ space minus space 1$ [ $because$ of (1)]
$rightwards double arrow space 2 space straight I space space equals space log space 2 open square brackets straight theta close square brackets subscript 0 superscript straight pi over 4 end superscript space equals space open parentheses straight pi over 4 minus 0 close parentheses space log space 2 space rightwards double arrow space space 2 space straight I space equals space straight pi over 4 log 2 space space rightwards double arrow space space straight I equals straight pi over 8 log space 2$