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Integrals

Short Answer Type

241.

If space straight f left parenthesis straight x right parenthesis space equals space integral subscript 0 superscript straight x straight t space sint space dt comma space write space the space value space of space straight f apostrophe left parenthesis straight x right parenthesis

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242.

If $integral subscript 0 superscript straight a fraction numerator 1 over denominator 4 plus straight x squared end fraction dx equals straight pi over 8$ , find the value of a.

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243.

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

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244.

Evaluate:
$integral fraction numerator straight x plus 2 over denominator square root of straight x squared plus 5 straight x plus 6 end root end fraction dx$

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Long Answer Type

245.
Evaluate:
$integral fraction numerator 1 over denominator sin to the power of 4 straight x plus sin squared xcos squared straight x plus cos to the power of 4 straight x end fraction dx$

We need to evaluate $integral fraction numerator dx over denominator sin to the power of 4 straight x plus sin squared xcos squared straight x plus cos to the power of 4 straight x end fraction$
$Let space straight I space equals integral fraction numerator dx over denominator sin to the power of 4 straight x plus sin squared xcos squared straight x plus cos to the power of 4 straight x end fraction$
Multiply the numerator and the denominator by $sec to the power of 4 straight x$ , we have

$straight I space equals space integral fraction numerator sec to the power of 4 xdx over denominator tan to the power of 4 straight x plus tan squared straight x plus 1 end fraction straight I space equals space integral fraction numerator sec squared straight x cross times sec squared xdx over denominator tan to the power of 4 straight x plus tan squared straight x plus 1 end fraction Now space substitute space straight t space equals space tanx semicolon space dt space equals space sec squared xdx Therefore comma space straight I space equals space integral fraction numerator left parenthesis 1 plus straight t squared right parenthesis dt over denominator straight t to the power of 4 plus straight t squared plus 1 end fraction straight I space equals space integral fraction numerator open parentheses 1 plus begin display style 1 over straight t squared end style close parentheses dt over denominator open parentheses straight t squared plus begin display style 1 over straight t squared end style plus 1 close parentheses end fraction straight I space equals integral fraction numerator open parentheses 1 plus begin display style 1 over straight t squared end style close parentheses dt over denominator open parentheses straight t squared plus begin display style 1 over straight t squared end style minus 2 plus 2 plus 1 close parentheses end fraction straight I space equals space integral fraction numerator open parentheses 1 plus begin display style 1 over straight t squared end style close parentheses dt over denominator open parentheses straight t minus begin display style 1 over straight t end style close parentheses squared plus 3 end fraction$
$Substitute space straight z equals straight t minus 1 over straight t semicolon space space dz space equals space open parentheses 1 plus 1 over straight t squared close parentheses dt straight I space equals space integral fraction numerator dz over denominator straight z squared plus 3 end fraction straight I space equals space integral fraction numerator dz over denominator straight z squared plus open parentheses square root of 3 close parentheses squared end fraction straight I space equals fraction numerator 1 over denominator square root of 3 end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator straight z over denominator square root of 3 end fraction close parentheses plus straight c straight I space equals fraction numerator 1 over denominator square root of 3 end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator straight t minus begin display style 1 over straight t end style over denominator square root of 3 end fraction close parentheses plus straight c$

$straight I equals space fraction numerator 1 over denominator square root of 3 end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator tanx minus begin display style 1 over tanx end style over denominator square root of 3 end fraction close parentheses plus straight c straight I space equals fraction numerator 1 over denominator square root of 3 end fraction tan to the power of negative 1 end exponent open parentheses fraction numerator tanx minus cotx over denominator square root of 3 end fraction close parentheses plus straight c$

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Short Answer Type

246.

Evaluate: $integral fraction numerator sin left parenthesis straight x minus straight a right parenthesis over denominator sin left parenthesis straight x plus straight a right parenthesis end fraction dx$

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247.

Evaluate:
$integral fraction numerator 5 straight x minus 2 over denominator left parenthesis 1 plus 2 straight x plus 3 straight x squared end fraction dx$

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248.

Evaluate:
$integral fraction numerator straight x squared over denominator left parenthesis straight x squared plus 4 right parenthesis left parenthesis straight x squared plus 9 right parenthesis end fraction dx$

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249.

Evaluate: $integral subscript 0 superscript 4 open parentheses open vertical bar straight x close vertical bar plus open vertical bar straight x minus 2 close vertical bar plus open vertical bar straight x minus 4 close vertical bar close parentheses dx$