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

Let I = $integral subscript 0 superscript straight pi fraction numerator straight x space tan space straight x over denominator sec space straight x plus cos space straight x end fraction dx$
$therefore space space space space straight I space equals space integral subscript 0 superscript straight pi fraction numerator left parenthesis straight pi minus straight x right parenthesis space tan left parenthesis straight pi minus straight x right parenthesis over denominator sec left parenthesis straight pi minus straight x right parenthesis plus cos left parenthesis straight pi minus straight x right parenthesis end fraction dx$ $open square brackets because space space integral subscript 0 superscript straight a straight f left parenthesis straight x right parenthesis space dx space equals space integral subscript 0 superscript straight a straight f left parenthesis straight a minus straight x right parenthesis space dx close square brackets$

$equals space integral subscript 0 superscript straight pi fraction numerator left parenthesis straight pi minus straight x right parenthesis space left parenthesis negative tanx right parenthesis over denominator negative secx minus cosx end fraction dx space equals space integral subscript 0 superscript straight pi fraction numerator left parenthesis straight pi minus straight x right parenthesis space tanx over denominator secx plus cosx end fraction dx$

$therefore space space space space space straight I space equals space integral subscript 0 superscript straight pi fraction numerator straight pi space tanx over denominator sec space straight x space plus space cos space straight x end fraction dx minus integral subscript 0 superscript straight pi fraction numerator straight x space tanx over denominator secx space plus space tanx end fraction dx$
Adding (1) and (2), we get
$2 space straight I space equals integral subscript 0 superscript straight pi fraction numerator straight pi space tanx over denominator secx plus cosx end fraction dx space equals space straight pi integral subscript 0 superscript straight pi fraction numerator tanx over denominator secx plus cosx end fraction dx space equals straight pi integral subscript 0 superscript straight pi fraction numerator begin display style sinx over cosx end style over denominator begin display style 1 over cosx end style plus cosx end fraction dx$
$therefore space space space space space space space space 2 straight I space equals space straight pi integral subscript 0 superscript straight pi fraction numerator sinxdx over denominator 1 plus cos squared straight x end fraction$

Put cos x = t, ∴ – sin x dx = dt, or sin x dx = – dt When x = 0, t = cos 0 = 1 When x = $straight pi$ , t = cos $straight pi$ = – 1
$therefore space space 2 space straight I space equals space minus straight pi integral subscript 1 superscript negative 1 end superscript fraction numerator dt over denominator 1 plus straight t squared end fraction space equals space straight pi integral subscript negative 1 end subscript superscript 1 fraction numerator dt over denominator 1 plus straight t squared end fraction space equals straight pi open square brackets tan to the power of negative 1 end exponent close square brackets subscript negative 1 end subscript superscript 1 space equals straight pi open square brackets tan to the power of negative 1 end exponent left parenthesis 1 right parenthesis space minus tan to the power of negative 1 end exponent left parenthesis negative 1 right parenthesis close square brackets space space space space space space space space space space space equals space straight pi open square brackets straight pi over 4 minus open parentheses negative straight pi over 4 close parentheses close square brackets space equals straight pi cross times straight pi over 2$

$therefore space space space 2 space straight I space equals space straight pi squared over 2 space space$ $rightwards double arrow space space space straight I space equals space straight pi over 4$

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