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

101 Views

205.
Show that:
$integral subscript 0 superscript straight pi fraction numerator straight x space tanx over denominator secx plus tanx end fraction dx space equals space straight pi open parentheses straight pi over 2 minus 1 close parentheses$

Let I = $integral subscript 0 superscript straight pi fraction numerator straight x space tanx over denominator secx plus tanx end fraction dx$ ...(1)
$therefore space space space space space straight I space equals space integral subscript 0 superscript straight pi fraction numerator left parenthesis straight pi minus straight x right parenthesis space tan left parenthesis straight pi minus straight x right parenthesis over denominator sec left parenthesis straight pi minus straight x right parenthesis plus tan left parenthesis straight pi minus straight x right parenthesis end fraction dx space space space space open square brackets because space integral subscript 0 superscript straight a straight f left parenthesis straight x right parenthesis space dx space equals space integral subscript 0 superscript straight a straight f left parenthesis straight a minus straight x right parenthesis space dx close square brackets$
$therefore space space space straight I space equals space integral subscript 0 superscript straight pi open square brackets fraction numerator left parenthesis straight pi minus straight x right parenthesis space left parenthesis negative tanx right parenthesis over denominator negative secx minus tanx end fraction close square brackets dx space space space rightwards double arrow space space straight I space equals space integral subscript 0 superscript straight pi fraction numerator left parenthesis straight pi minus straight x right parenthesis space tanx over denominator secx plus tanx end fraction dx$
$rightwards double arrow space space space space space straight I space equals straight pi integral subscript 0 superscript straight pi fraction numerator tanx over denominator secx plus tanx end fraction dx space minus space integral subscript 0 superscript straight pi fraction numerator straight x space tanx over denominator secx plus tanx end fraction dx$

$rightwards double arrow space space straight I space equals space straight pi integral subscript 0 superscript straight pi over 2 end superscript fraction numerator tan space straight x over denominator secx plus tanx end fraction cross times fraction numerator secx minus tanx over denominator secx minus tanx end fraction dx space minus space 1$ [ $because space of space left parenthesis 1 right parenthesis$ ]

$rightwards double arrow space space space space 2 space straight I space equals space straight pi space integral subscript 0 superscript straight pi over 2 end superscript fraction numerator tanx space secx minus tan squared straight x over denominator sec squared straight x minus tan squared straight x end fraction dx space space space rightwards double arrow space space space 2 space straight I space space equals space straight pi integral subscript 0 superscript straight pi fraction numerator secx space tanx minus left parenthesis sec squared straight x minus 1 right parenthesis over denominator 1 end fraction dx rightwards double arrow space space space 2 straight I space equals space straight pi integral subscript 0 superscript straight pi left parenthesis secx space tanx space minus sec squared straight x plus 1 right parenthesis space dx space equals space straight pi left square bracket secx minus tanx plus straight x right square bracket subscript 0 superscript straight pi$
$equals space straight pi open square brackets secπ minus tanπ plus straight pi right parenthesis space minus space left parenthesis sec space 0 space minus tan space 0 space plus 0 right parenthesis close square brackets equals straight pi open square brackets left parenthesis negative 1 minus 0 plus straight pi right parenthesis space minus space left parenthesis 1 minus 0 plus 0 right parenthesis close square brackets$

$rightwards double arrow space space 2 space straight I space equals space straight pi left parenthesis straight pi minus 2 right parenthesis rightwards double arrow space space space space space straight I space equals space straight pi open parentheses straight pi over 2 minus 1 close parentheses$

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