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

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

Let I = $integral subscript 0 superscript straight pi fraction numerator straight x space sinx over denominator 1 plus cos squared straight x end fraction dx$
$therefore space space space straight I space equals space integral subscript 0 superscript straight pi fraction numerator left parenthesis straight pi minus straight x right parenthesis space sin left parenthesis straight pi minus straight x right parenthesis over denominator 1 plus cos squared left parenthesis straight pi minus straight x right parenthesis end fraction dx space space space space space space space space space space space space space open square brackets because space space integral subscript 0 superscript straight a straight f left parenthesis straight x right parenthesis space dx space equals space integral subscript 0 superscript straight a straight f left parenthesis straight a minus straight x right parenthesis space dx close square brackets$
$equals space integral subscript 0 superscript straight pi fraction numerator left parenthesis straight pi minus straight x right parenthesis space sinx over denominator 1 plus cos squared straight x end fraction dx space equals space straight pi integral subscript 0 superscript straight pi fraction numerator sinx over denominator 1 plus cos squared straight x end fraction dx minus integral subscript 0 superscript straight pi fraction numerator straight x space sinx over denominator 1 plus cos squared straight x end fraction dx$
$therefore space space space straight I space equals space straight pi integral subscript 0 superscript straight pi fraction numerator sinx over denominator 1 plus cos squared straight x end fraction dx space minus space 1 therefore space space 2 space straight I space equals space straight pi integral subscript 0 superscript straight pi fraction numerator sin space straight x space over denominator 1 plus cos squared straight x end fraction dx$
Put cos x = t. ∴ sin x dx = – dt When x = 0, t = cos 0 = 1 When x = $straight pi$ , t = cos $straight pi$ = – 1
$therefore space space space 2 space straight I space equals space straight pi integral subscript 1 superscript negative 1 end superscript fraction numerator dt over denominator 1 plus straight t squared end fraction space equals space straight pi open square brackets tan to the power of negative 1 end exponent straight t close square brackets subscript 1 superscript negative 1 end superscript space space space space space space space space space space space space space equals negative straight pi open square brackets tan to the power of negative 1 end exponent left parenthesis negative 1 right parenthesis space minus space tan to the power of negative 1 end exponent 1 close square brackets space equals space minus straight pi open square brackets negative straight pi over 4 minus straight pi over 4 close square brackets therefore space space 2 space straight I space equals space straight pi squared over 2 space space rightwards double arrow space space space space straight I space equals space straight pi squared over 4$