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Integrals

Long Answer Type

211.
Evaluate:
$integral subscript 0 superscript straight pi fraction numerator straight x over denominator 1 plus sin squared straight x end fraction dx.$

Let I = $integral subscript 0 superscript straight pi fraction numerator straight x over denominator 1 plus sin squared straight x end fraction dx$ ...(1)
$therefore space space space space space space straight I space equals space integral subscript 0 superscript straight pi fraction numerator straight pi minus straight x over denominator 1 plus sin squared left parenthesis straight pi minus straight x right parenthesis end fraction dx space space space space space rightwards double arrow space space space space straight I space equals space integral subscript 0 superscript straight pi fraction numerator straight pi minus straight x over denominator 1 plus sin squared straight x end fraction dx$
$rightwards double arrow space space space space space space space straight I space equals space straight pi integral subscript 0 superscript straight pi fraction numerator 1 over denominator 1 plus sin squared straight x end fraction dx space space minus integral subscript 0 superscript straight pi fraction numerator straight x over denominator 1 plus sin squared straight x end fraction dx$
$therefore space space space space space straight I space equals space straight pi integral subscript 0 superscript straight pi fraction numerator 1 over denominator 1 plus sin squared straight x end fraction dx minus 1 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because space space of space left parenthesis 1 right parenthesis close square brackets$

$therefore space space space space space 2 space space straight I space equals space straight pi integral subscript 0 superscript straight pi fraction numerator 1 over denominator 1 plus sin squared straight x end fraction dx space equals space 2 straight pi space integral subscript 0 superscript straight pi over 2 end superscript fraction numerator 1 over denominator 1 plus sin squared straight x end fraction dx$

$equals space 2 straight pi integral subscript 0 superscript straight pi over 2 end superscript fraction numerator 1 over denominator sin squared straight x plus cos squared straight x plus sin squared straight x end fraction dx space equals space 2 straight pi integral subscript 0 superscript straight pi over 2 end superscript fraction numerator 1 over denominator 2 sin squared straight x plus cos squared straight x end fraction dx equals space 2 straight pi integral subscript 0 superscript straight pi over 2 end superscript fraction numerator begin display style fraction numerator 1 over denominator cos squared straight x end fraction end style over denominator begin display style fraction numerator 2 sin squared straight x over denominator cos squared straight x end fraction end style plus begin display style fraction numerator cos squared straight x over denominator cos squared straight x end fraction end style end fraction dx space equals space 2 straight pi integral subscript 0 superscript straight pi over 2 end superscript fraction numerator sec squared straight x over denominator 2 space tan squared straight x plus 1 end fraction dx$
Put tan x = t, ∴ sec ² x dx = dt When x = 0, t = tan 0 = 0
When $straight x space equals space straight pi over 2 comma space space straight t space equals space tan straight pi over 2 space rightwards arrow space space infinity$
$therefore space space space space space space space space space space straight I space equals space straight pi integral subscript 0 superscript infinity fraction numerator 1 over denominator 2 straight t squared plus 1 end fraction dt space equals space straight pi over 2 integral subscript 0 superscript infinity fraction numerator 1 over denominator straight t squared plus begin display style 1 half end style end fraction dt space equals space straight pi over 2 integral subscript 0 superscript infinity fraction numerator 1 over denominator straight t squared plus open parentheses begin display style fraction numerator 1 over denominator square root of 2 end fraction end style close parentheses squared end fraction dt$
$equals space straight pi over 2 cross times fraction numerator 1 over denominator begin display style fraction numerator 1 over denominator square root of 2 end fraction end style end fraction open square brackets tan to the power of negative 1 end exponent open parentheses fraction numerator straight t over denominator begin display style fraction numerator 1 over denominator square root of 2 end fraction end style end fraction close parentheses close square brackets subscript 0 superscript infinity space equals space fraction numerator straight pi over denominator square root of 2 end fraction open square brackets tan to the power of negative 1 end exponent open parentheses square root of 2 straight t close parentheses close square brackets subscript 0 superscript infinity equals space fraction numerator straight pi over denominator square root of 2 end fraction left square bracket tan to the power of negative 1 end exponent infinity space minus space tan to the power of negative 1 end exponent 0 right square bracket space equals space fraction numerator straight pi over denominator square root of 2 end fraction open parentheses fraction numerator straight pi over denominator square root of 2 end fraction minus 0 close parentheses space equals space fraction numerator straight pi over denominator square root of 2 end fraction cross times straight pi over 2 space equals fraction numerator straight pi squared over denominator 2 square root of 2 end fraction$

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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

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

131 Views

217.

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

231 Views

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