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Integrals

Short Answer Type

161.
Show that:
$integral subscript 0 superscript straight pi over 2 end superscript fraction numerator cos space straight x over denominator sin space straight x space plus space cos space straight x end fraction dx$

Let I = $integral subscript 0 superscript straight pi over 2 end superscript fraction numerator cos space straight x over denominator sin space straight x plus space cos space straight x end fraction dx$ ...(1)
Then $straight I space equals space integral subscript 0 superscript straight pi over 2 end superscript fraction numerator cos space open parentheses begin display style straight pi over 2 end style minus straight x close parentheses over denominator sin open parentheses begin display style straight pi over 2 end style minus straight x close parentheses plus cos open parentheses begin display style straight pi over 2 end style minus straight x close parentheses end fraction dx space space space space open square brackets because space integral subscript 0 superscript straight a straight f left parenthesis straight x right parenthesis space dx space equals space integral subscript 0 superscript straight a straight f left parenthesis straight a minus straight x right parenthesis space dx close square brackets$
$therefore space space straight I space equals space integral subscript 0 superscript straight pi over 2 end superscript fraction numerator sin space straight x over denominator cosx plus sinx end fraction$ ...(2)
Adding (1) and (2), we get.
$2 straight I space equals space integral subscript 0 superscript straight pi over 2 end superscript open parentheses fraction numerator cos space straight x over denominator sin space straight x space plus space cos space straight x space end fraction plus fraction numerator sin space straight x over denominator sin space straight x plus space cos space straight x end fraction close parentheses dx space equals space integral subscript 0 superscript straight pi over 2 end superscript fraction numerator cosx plus sinx over denominator sinx plus cosx end fraction dx space equals integral subscript 0 superscript straight pi over 2 end superscript 1 space dx$

$equals space open square brackets straight x close square brackets subscript 0 superscript straight pi over 2 end superscript space equals space straight pi over 2 minus 0 space equals space straight pi over 2$ .

$therefore space space space space space space 2 space space straight I space space equals space space straight pi over 2 space space space space space space space space space rightwards double arrow space space space space straight I space equals space straight pi over 4$

81 Views

Long Answer Type

162.

Evaluate $integral subscript straight pi over 6 end subscript superscript straight pi over 3 end superscript fraction numerator dx over denominator 1 plus square root of tanx end fraction$

92 Views

163.

Evaluate $integral subscript straight pi over 6 end subscript superscript straight pi over 3 end superscript fraction numerator square root of sin space straight x end root over denominator square root of sin space straight x end root plus square root of cos space straight x end root end fraction dx$

97 Views

Short Answer Type

164.

Evaluate $integral subscript 0 superscript straight pi space sin squared straight x space cos cubed straight x space dx.$

115 Views

165.

Evaluate $integral subscript negative straight pi over 4 end subscript superscript straight pi over 4 end superscript space sin squared straight x space dx$

105 Views

166.

Evaluate $integral subscript negative straight pi over 4 end subscript superscript straight pi over 4 end superscript cos squared xdx$

116 Views

167.

By using the properties of definite integrals, evaluate
$integral subscript negative straight pi over 2 end subscript superscript straight pi over 2 end superscript cosx space dx$

98 Views

168.

Evaluate $integral subscript negative 1 end subscript superscript 1 sin to the power of 5 straight x space cos to the power of 4 straight x space dx$

108 Views

169.

Evaluate $integral subscript negative 1 end subscript superscript 1 space straight x to the power of 17 space cos to the power of 4 straight x space dx.$

85 Views

170.

Evaluate $integral subscript negative 1 end subscript superscript 1 space log space open parentheses fraction numerator 2 plus straight x over denominator 2 minus straight x end fraction close parentheses dx.$