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Integrals

Short Answer Type

121.

Evaluate: $integral subscript 0 superscript straight pi over 4 end superscript space fraction numerator sinx space cosx over denominator cos squared straight x plus sin to the power of 4 straight x end fraction dx.$

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122.

Evaluate the following definite integral:
$integral subscript 0 superscript straight pi over 4 end superscript fraction numerator sinx space cosx over denominator cos to the power of 4 straight x space plus space sin to the power of 4 straight x end fraction dx$

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123.

Evaluate the following definite integral:
$integral subscript straight pi over 6 end subscript superscript straight pi over 3 end superscript fraction numerator sinx plus cosx over denominator square root of sin space 2 straight x end root end fraction dx.$

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124.

Evaluate the following definite integral:
$integral subscript 0 superscript straight pi over 4 end superscript fraction numerator sinx plus cosx over denominator 9 plus 16 sin 2 straight x end fraction dx$

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125.

Evaluate the following definite integral:
$integral subscript 0 superscript straight pi over 2 end superscript sin space 2 straight x space tan to the power of negative 1 end exponent left parenthesis sin space straight x right parenthesis space dx.$

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126.

Prove that $integral subscript 0 superscript straight pi fraction numerator dx over denominator 5 plus 3 cosx end fraction space equals space straight pi over 4$

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127.

Evaluate $integral subscript 0 superscript straight pi fraction numerator dx over denominator 5 plus 4 space cosx end fraction.$

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128.
Evaluate the following:
$integral subscript 0 superscript straight pi over 2 end superscript fraction numerator dx over denominator 5 plus 4 space sinx end fraction$

Let I = $integral subscript 0 superscript straight pi over 2 end superscript fraction numerator dx over denominator 5 plus 4 space sinx end fraction$
Put $tan space straight x over 2 space equals space straight t space space space space space space space space or space space space straight x over 2 space equals space tan to the power of negative 1 end exponent straight t comma space space space space space therefore space space straight x space equals space 2 space tan to the power of negative 1 end exponent straight t space space space space space space space space space space space rightwards double arrow space space dx space equals space fraction numerator 2 over denominator 1 plus straight t squared end fraction$
$sinx space equals space fraction numerator 2 tan begin display style straight x over 2 end style over denominator 1 plus tan squared begin display style straight x over 2 end style end fraction space equals space fraction numerator 2 straight t over denominator 1 plus straight t squared end fraction$
When x = 0, t = tan 0 = 0
When $straight x space equals space straight pi over 2 comma space space straight t space equals space tan straight pi over 4 space equals space 1$
$therefore$ $straight I space equals space integral subscript 0 superscript 1 fraction numerator begin display style fraction numerator 2 space dt over denominator 1 plus straight t squared end fraction end style over denominator 5 plus 4 open parentheses begin display style fraction numerator 2 straight t over denominator 1 plus straight t squared end fraction end style close parentheses end fraction space equals space integral subscript 0 superscript 1 fraction numerator 2 space dt over denominator 5 plus 5 straight t squared plus 8 straight t end fraction space equals space 2 over 5 integral subscript 0 superscript 1 fraction numerator dt over denominator straight t squared plus begin display style 8 over 5 end style straight t plus 1 end fraction 2 over 5$ $equals space integral subscript 0 superscript 1 fraction numerator dt over denominator open parentheses straight r squared plus begin display style 8 over 5 end style straight t plus begin display style 16 over 25 end style close parentheses plus begin display style 9 over 25 end style end fraction space equals space 2 over 5 integral subscript 0 superscript 1 fraction numerator dt over denominator open parentheses straight t plus begin display style 4 over 5 end style close parentheses squared plus open parentheses begin display style 3 over 5 end style close parentheses squared end fraction space equals 2 over 5. fraction numerator 1 over denominator begin display style 3 over 5 end style end fraction open square brackets tan to the power of negative 1 end exponent open parentheses fraction numerator straight t plus begin display style 4 over 5 end style over denominator begin display style 3 over 5 end style end fraction close parentheses close square brackets subscript 0 superscript 1$

$equals space 2 over 3 open square brackets tan to the power of negative 1 end exponent open parentheses fraction numerator 5 straight t plus 4 over denominator 3 end fraction close parentheses close square brackets subscript 0 superscript 1 space equals space 2 over 3 open square brackets tan to the power of negative 1 end exponent space 3 space minus space tan to the power of negative 1 end exponent 4 over 3 close square brackets space equals space 2 over 3 tan to the power of negative 1 end exponent open square brackets fraction numerator 3 minus begin display style 4 over 3 end style over denominator 1 plus 3 cross times begin display style 4 over 3 end style end fraction close square brackets space space space space space space space space space space space space space space open square brackets because space space tan to the power of negative 1 end exponent straight a minus tan to the power of negative 1 end exponent straight b space equals space tan to the power of negative 1 end exponent open parentheses fraction numerator straight a minus straight b over denominator 1 plus straight a space straight b end fraction close parentheses close square brackets equals space 2 over 3 tan to the power of negative 1 end exponent open parentheses fraction numerator begin display style 5 over 3 end style over denominator 5 end fraction close parentheses space equals space 2 over 3 tan to the power of negative 1 end exponent open parentheses 1 third close parentheses$