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Integrals

Long Answer Type

211.

Evaluate:
$integral subscript 0 superscript straight pi fraction numerator straight x over denominator 1 plus sin squared straight x end fraction dx.$

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Short Answer Type

212. By using the properties of definite integrals, evaluate the following:
$integral subscript 0 superscript straight pi log space left parenthesis 1 plus cosx right parenthesis space dx$

Let I = $integral subscript 0 superscript straight pi log left parenthesis 1 plus cosx right parenthesis space dx$ ...(1)
Then I = $integral subscript 0 superscript straight pi log left parenthesis 1 plus cos stack straight pi minus straight x with bar on top right parenthesis space dx$
$therefore space space space space straight I space equals space integral subscript 0 superscript straight pi log left parenthesis 1 minus cosx right parenthesis space dx$ ...(2)
Adding (1) and (2), we get,
$2 space straight I space equals space integral subscript 0 superscript straight pi open square brackets log left parenthesis 1 plus cosx right parenthesis plus log left parenthesis 1 minus cosx right parenthesis close square brackets dx space equals space integral subscript 0 superscript straight pi log open square brackets left parenthesis 1 plus cosx right parenthesis space left parenthesis 1 minus cosx right parenthesis close square brackets dx$
$equals space integral subscript 0 superscript straight pi log left parenthesis 1 minus cos squared straight x right parenthesis space dx space equals space integral subscript 0 superscript straight pi log space left parenthesis sin squared straight x right parenthesis space dx space equals space 2 integral subscript 0 superscript straight pi log space sinx space dx$
$equals space 2.2 space integral subscript 0 superscript straight pi over 2 end superscript log space sinx space dx$ $open square brackets because space space log space sin left parenthesis straight pi minus straight x right parenthesis space equals space log space sinx close square brackets$
$equals space 4 open parentheses negative straight pi over 2 log 2 close parentheses space equals space minus 2 space straight pi space log space 2$
$therefore space space straight I space equals space minus straight pi space log space 2$

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213. By using the properties of definite integrals, evaluate the following:

integral subscript 0 superscript straight pi over 2 end superscript fraction numerator straight x space sinx space cosx over denominator sin to the power of 4 straight x plus cos to the power of 4 straight x end fraction dx

153 Views

Long Answer Type

214. By using the properties of definite integrals, evaluate the following:

integral subscript 0 superscript straight pi fraction numerator straight x space dx over denominator straight a squared space cos squared straight x plus straight b squared space sin squared straight x end fraction

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Short Answer Type

215.

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

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216.

Show that:
$integral subscript 0 superscript straight a fraction numerator 1 over denominator straight x plus square root of straight a squared minus straight x squared end root end fraction dx space equals space straight pi over 4$

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217.

Evaluate $integral subscript 1 superscript 4 straight f left parenthesis straight x right parenthesis space dx space space where space straight f left parenthesis straight x right parenthesis space equals space open vertical bar straight x minus 1 close vertical bar space plus space open vertical bar straight x minus 2 close vertical bar space plus space open vertical bar straight x minus 3 close vertical bar$

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218.

Evaluate: $integral subscript negative 5 end subscript superscript 0 space straight f left parenthesis straight x right parenthesis space dx comma space space where space straight f left parenthesis straight x right parenthesis space equals space open vertical bar straight x close vertical bar plus open vertical bar straight x plus 2 close vertical bar plus open vertical bar straight x plus 5 close vertical bar.$

149 Views

219.

Evaluate: $integral subscript negative 5 end subscript superscript 0 straight f left parenthesis straight x right parenthesis space dx comma space space space where space straight f left parenthesis straight x right parenthesis space equals space open vertical bar straight x close vertical bar plus open vertical bar straight x plus 3 close vertical bar plus open vertical bar straight x plus 6 close vertical bar.$

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220.

If [ x ] stands for integral part of x, then show that $integral subscript 0 superscript 1 left square bracket 5 space straight x right square bracket space dx space equals space 2.$

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