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Integrals

Short Answer Type

191.
Show that:
$integral subscript straight pi over 4 end subscript superscript straight pi over 4 end superscript open vertical bar sin space straight x close vertical bar dx space equals space 2 minus square root of 2$

Let I = $integral subscript negative straight pi over 4 end subscript superscript straight pi over 4 end superscript space open vertical bar sin space straight x close vertical bar space dx$
For $negative straight pi over 4 less or equal than straight x less or equal than 0 comma space space space open vertical bar sin space straight x close vertical bar space equals space minus sin space straight x space$
and for $0 less or equal than straight x less or equal than straight pi over 4 comma space space open vertical bar sin space straight x close vertical bar space equals space sin space straight x$
$therefore space space space straight I space equals space integral subscript negative straight pi over 4 end subscript superscript 0 open vertical bar sin space straight x close vertical bar dx plus integral subscript 0 superscript straight pi over 4 end superscript open vertical bar sin space straight x close vertical bar dx space equals space minus integral subscript negative straight pi over 4 end subscript superscript 0 sinx space dx plus integral subscript 0 superscript straight pi over 4 end superscript sinx space dx$
$equals space open square brackets cos space straight x close square brackets subscript negative straight pi over 4 end subscript superscript 0 space minus space open square brackets cos space straight x close square brackets subscript 0 superscript straight pi over 4 end superscript space equals space open square brackets cos space 0 space minus space cos space open parentheses negative straight pi over 4 close parentheses close square brackets space minus space open square brackets cos space straight pi over 4 minus cos space 0 close square brackets equals space open parentheses 1 minus fraction numerator 1 over denominator square root of 2 end fraction close parentheses space minus space open square brackets fraction numerator 1 over denominator square root of 2 end fraction minus 1 close square brackets space equals space 2 minus fraction numerator 2 over denominator square root of 2 end fraction space equals space 2 minus square root of 2$

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