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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

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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

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

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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 $integral subscript 0 superscript 1 left square bracket 5 space straight x right square bracket space dx space equals space 2.$

Since $open square brackets 5 space straight x close square brackets$ has integral values at x = 0, $1 fifth comma space 2 over 5 comma space 3 over 5 comma space 4 over 5 comma space 1$

$therefore space space space space left square bracket 5 space straight x right square bracket$ is discontinuous at x = 0, $1 fifth comma space 2 over 5 comma space 3 over 5 comma space 4 over 5 comma space 1$

$therefore space space straight I space equals space integral subscript 0 superscript 1 left square bracket space 5 space straight x right square bracket space dx space equals space integral subscript 0 superscript 1 fifth end superscript left square bracket space 5 space straight x right square bracket space dx space plus space integral subscript 1 fifth end subscript superscript 2 over 5 end superscript left square bracket space 5 space straight x right square bracket space dx space plus space integral subscript 2 over 5 end subscript superscript 3 over 5 end superscript left square bracket 5 space straight x right square bracket space dx plus integral subscript 3 over 5 end subscript superscript 4 over 5 end superscript left square bracket 5 space straight x right square bracket space dx space plus space integral subscript 4 over 5 end subscript superscript 1 left square bracket 5 space straight x right square bracket space dx$

$equals space integral subscript 0 superscript 1 fifth end superscript left parenthesis 0 right parenthesis space dx space plus space integral subscript 1 fifth end subscript superscript 2 over 5 end superscript left parenthesis 1 right parenthesis space dx space plus space integral subscript 2 over 5 end subscript superscript 3 over 5 end superscript left parenthesis 2 right parenthesis space dx space plus space integral subscript 3 over 5 end subscript superscript 4 over 5 end superscript left parenthesis 3 right parenthesis space dx space plus space integral subscript 4 over 5 end subscript superscript 1 left parenthesis 4 right parenthesis space dx$

$equals 0 plus open square brackets straight x close square brackets subscript 1 fifth end subscript superscript 2 over 5 end superscript plus 2 open square brackets straight x close square brackets subscript 2 over 5 end subscript superscript 3 over 5 end superscript plus 3 open square brackets straight x close square brackets subscript 3 over 5 end subscript superscript 4 over 5 end superscript plus space 4 open square brackets straight x close square brackets subscript 4 over 5 end subscript superscript 1 equals space 0 plus open parentheses 2 over 5 minus 1 fifth close parentheses plus 2 open parentheses 3 over 5 minus 2 over 5 close parentheses plus 3 open parentheses 4 over 5 minus 3 over 5 close parentheses plus 4 open parentheses 1 minus 4 over 5 close parentheses equals space 0 plus 1 fifth plus 2 over 5 plus 3 over 5 plus 4 over 5 equals space space 2$