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Integrals

Short Answer Type

161.

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

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Long Answer Type

162.

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

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163.
Evaluate $integral subscript straight pi over 6 end subscript superscript straight pi over 3 end superscript fraction numerator square root of sin space straight x end root over denominator square root of sin space straight x end root plus square root of cos space straight x end root end fraction dx$

Let I = $integral subscript straight pi over 6 end subscript superscript straight pi over 3 end superscript fraction numerator square root of sin space straight x end root over denominator square root of sin space straight x end root plus square root of cos space straight x end root end fraction dx$ ...(1)
Then I = $integral subscript straight pi over 6 end subscript superscript straight pi over 3 end superscript fraction numerator square root of sin space open parentheses begin display style straight pi over 3 end style plus begin display style straight pi over 6 end style minus straight x close parentheses end root over denominator square root of sin space open parentheses begin display style straight pi over 3 end style plus begin display style straight pi over 6 end style minus straight x close parentheses end root plus square root of cos space open parentheses begin display style straight pi over 3 end style plus begin display style straight pi over 6 end style minus straight x close parentheses end root end fraction dx$
   $open square brackets because space space integral subscript straight a superscript straight b straight f left parenthesis straight x right parenthesis space dx space equals space integral subscript straight a superscript straight b straight f left parenthesis straight a plus straight b minus straight x right parenthesis space dx close square brackets$

$equals space integral subscript straight pi over 6 end subscript superscript straight pi over 3 end superscript fraction numerator square root of sin space open parentheses begin display style straight pi over 2 end style minus straight x close parentheses end root over denominator square root of sin space open parentheses begin display style straight pi over 2 end style minus straight x close parentheses end root plus square root of cos space open parentheses begin display style straight pi over 2 end style minus straight x close parentheses end root end fraction dx$
$therefore space space space straight I space equals space integral subscript straight pi over 6 end subscript superscript straight pi over 3 end superscript fraction numerator square root of cos space straight x end root over denominator square root of cos space straight x end root plus square root of sin space straight x end root end fraction dx$ ...(2)

Adding (1) and (2), we get,
   $2 space straight I space equals space integral subscript straight pi over 6 end subscript superscript straight pi over 3 end superscript open square brackets fraction numerator square root of sin space straight x end root over denominator square root of sin space straight x end root plus square root of cos space straight x end root end fraction plus negative fraction numerator square root of cos space straight x end root over denominator square root of sin space straight x end root plus square root of cos space straight x end root end fraction close square brackets dx$

   $equals space integral subscript straight pi over 6 end subscript superscript straight pi over 3 end superscript fraction numerator square root of sin space straight x end root plus square root of cos space straight x end root over denominator square root of sin space straight x end root plus square root of cos space straight x end root end fraction dx space equals space integral subscript straight pi over 6 end subscript superscript straight pi over 3 end superscript 1 space dx space equals space open square brackets straight x close square brackets subscript straight pi divided by 6 end subscript superscript straight pi divided by 3 end superscript space equals space straight pi over 3 minus straight pi over 6 space equals fraction numerator 2 straight pi minus straight pi over denominator 6 end fraction$

$therefore space space 2 space straight I space equals space straight pi over 6 space space space rightwards double arrow space space straight I space equals space straight pi over 12$

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

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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
$integral subscript negative straight pi over 2 end subscript superscript straight pi over 2 end superscript cosx space dx$

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168.

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

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

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Previous Year Papers

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Short Answer Type

Long Answer Type

163. Evaluate

Short Answer Type