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Integrals

Multiple Choice Questions

141.

If $straight f left parenthesis straight x right parenthesis space equals space integral subscript 0 superscript straight x space straight t space sin space straight t space dt comma space space then space straight f apostrophe left parenthesis straight x right parenthesis space is$

cosx + x sin x
x sinx
x cosx
x cosx

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Short Answer Type

142.

Evaluate $integral subscript 0 superscript straight pi divided by 2 end superscript space sin squared straight x space dx$

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143. By using the properties of definite integrals, evaluate the following:

integral subscript 0 superscript straight pi over 2 end superscript space cos squared straight x space dx

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144. By using the properties of definite integrals, evaluate the following:

integral subscript 0 superscript straight pi over 2 end superscript fraction numerator cos to the power of 5 straight x over denominator sin to the power of 5 straight x plus cos to the power of 5 straight x end fraction dx

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145. By using the properties of definite integrals, evaluate the following:

integral subscript 0 superscript straight pi over 2 end superscript fraction numerator square root of cosx over denominator square root of cosx plus square root of sinx end fraction dx

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Long Answer Type

146.

Prove that: $integral subscript 0 superscript straight pi over 2 end superscript fraction numerator dx over denominator 1 plus tanx end fraction space equals space straight pi over 4.$

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147.

Evaluate the following:
$integral subscript 0 superscript straight pi over 2 end superscript fraction numerator dx over denominator 1 plus tan cubed straight x end fraction.$

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148.
Evaluate: $integral subscript 0 superscript straight pi over 2 end superscript fraction numerator 1 over denominator 1 plus tan to the power of 5 straight x end fraction dx$

Let I = $straight I space equals integral subscript 0 superscript straight pi over 2 end superscript fraction numerator 1 over denominator 1 plus tan to the power of 5 straight x end fraction dx space equals space integral subscript 0 superscript straight pi over 2 end superscript fraction numerator 1 over denominator 1 plus begin display style fraction numerator sin to the power of 5 straight x over denominator cos to the power of 5 straight x end fraction end style end fraction dx$
$therefore space space space space space space straight I space equals space integral subscript 0 superscript straight pi over 2 end superscript fraction numerator cos to the power of 5 straight x over denominator cos to the power of 5 straight x plus sin to the power of 5 straight x end fraction dx$ ...(1)

$therefore space space space space straight I space equals space integral subscript 0 superscript straight pi over 2 end superscript fraction numerator cos to the power of 5 open parentheses begin display style straight pi over 2 end style minus straight x close parentheses over denominator cos to the power of 5 open parentheses begin display style straight pi over 2 end style minus straight x close parentheses plus sin to the power of 5 open parentheses begin display style straight pi over 2 end style minus straight x close parentheses end fraction space open square brackets because space space integral subscript 0 superscript straight a straight f left parenthesis straight x right parenthesis dx space equals space integral subscript 0 superscript straight a straight f left parenthesis straight a minus straight x right parenthesis space dx close square brackets$
...(2)
Adding (1) and (2), we get,
$therefore space space space space space space space space space space 2 space straight I space equals space integral subscript 0 superscript straight pi over 2 end superscript open parentheses fraction numerator cos to the power of 5 straight x over denominator sin to the power of 5 straight x plus cos to the power of 5 straight x end fraction plus fraction numerator sin to the power of 5 straight x over denominator sin to the power of 5 straight x plus cos to the power of 5 straight x end fraction close parentheses dx$

$space equals space integral subscript 0 superscript straight pi over 2 end superscript fraction numerator sin to the power of 5 straight x plus cos to the power of 5 straight x over denominator sin to the power of 5 straight x plus cos to the power of 5 straight x end fraction dx space equals space integral subscript 0 superscript straight pi over 2 end superscript space 1 space dx space space equals space open square brackets straight x close square brackets subscript 0 superscript straight pi over 2 end superscript space equals space straight pi over 2 minus 0 therefore space space space space space 2 space space space straight I space space equals space straight pi over 2 space space space space space rightwards double arrow space space space space straight I space equals space straight pi over 4$

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149.

Evaluate $integral subscript 0 superscript straight pi over 2 end superscript fraction numerator 1 over denominator 1 plus cot cubed straight x end fraction dx$

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Short Answer Type

150.

Show that:
$integral subscript 0 superscript straight pi over 2 end superscript fraction numerator sin space straight x over denominator sin space straight x plus space cos space straight x end fraction dx space equals space straight pi over 4$