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

Let I = $integral subscript 0 superscript straight pi over 2 end superscript fraction numerator straight x space tanx over denominator secx space cosecx end fraction dx$ ...(1)
$therefore space space space space space space space straight I space equals space integral subscript 0 superscript straight pi over 2 end superscript fraction numerator left parenthesis straight pi minus straight x right parenthesis space tan left parenthesis straight pi minus straight x right parenthesis over denominator sec left parenthesis straight pi minus straight x right parenthesis space cosec left parenthesis straight pi minus straight x right parenthesis end fraction dx$ $open square brackets because space space integral subscript 0 superscript straight a straight f left parenthesis straight x right parenthesis space dx space equals space integral subscript 0 superscript straight a straight f left parenthesis straight a minus straight x right parenthesis space dx close square brackets$
$therefore space space space space space space space space straight I space equals space integral subscript 0 superscript straight pi over 2 end superscript fraction numerator left parenthesis straight pi minus straight x right parenthesis space left parenthesis negative tan space straight x right parenthesis over denominator left parenthesis negative sec space straight x right parenthesis space cosecx end fraction dx$
$straight I space equals straight pi integral subscript 0 superscript straight pi fraction numerator tan space straight x over denominator sec space straight x space cosec space straight x end fraction dx space space minus space integral subscript 0 superscript straight pi fraction numerator straight x space tanx over denominator sec space straight x space cosec space straight x end fraction dx$
$rightwards double arrow space space space space straight I space equals space straight pi integral subscript 0 superscript straight pi over 2 end superscript fraction numerator begin display style sinx over cosx end style over denominator begin display style 1 over cosx end style. begin display style 1 over sinx end style end fraction dx minus 1$ $open square brackets because space of space left parenthesis 1 right parenthesis close square brackets$
$rightwards double arrow space space space space 2 space straight I space equals space straight pi integral subscript 0 superscript straight pi over 2 end superscript sin squared space xdx space equals space straight pi over 2 integral subscript 0 superscript straight pi over 2 end superscript 2 space sin squared straight x space dx rightwards double arrow space space space space 2 space straight I space equals space straight pi over 2 integral subscript 0 superscript straight pi left parenthesis 1 minus cos space 2 straight x right parenthesis space dx space equals space straight pi over 2 open square brackets straight x minus fraction numerator sin space 2 straight x over denominator 2 end fraction close square brackets subscript 0 superscript straight pi therefore space space space space space 2 space straight I space equals space straight pi over 2 open square brackets open parentheses straight pi minus 1 half sin space 2 straight pi close parentheses space minus space open parentheses 0 minus 1 half sin space 0 close parentheses close square brackets therefore space space space 2 space straight I space equals space straight pi over 2 open square brackets open parentheses straight pi over 2 minus 1 half cross times 0 close parentheses space minus space left parenthesis 0 minus 0 right parenthesis close square brackets space space space space rightwards double arrow space space space 2 space straight I space equals space straight pi squared over 2 space space rightwards double arrow space space straight I space equals straight pi squared over 4$
$therefore space space space space straight I space equals space integral subscript 0 superscript straight pi over 2 end superscript fraction numerator left parenthesis straight pi minus straight x right parenthesis space left parenthesis negative tan space straight x right parenthesis over denominator left parenthesis negative sec space straight x right parenthesis space cosecx end fraction dx$

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