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

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129. Evaluate the following:

integral subscript 0 superscript straight pi fraction numerator 1 over denominator 5 plus 2 cosx end fraction dx

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130. Evaluate the following:
$integral subscript 0 superscript straight pi fraction numerator 1 over denominator 6 minus cosx end fraction dx$

Let I = $integral subscript 0 superscript straight pi fraction numerator 1 over denominator 6 minus cosx end fraction dx$
Put $tan space straight x over 2 space equals space straight t comma space space space or space space straight x over 2 space equals space tan to the power of negative 1 end exponent straight t comma space space space or space space straight x space equals space 2 space tan to the power of negative 1 end exponent space straight t space space rightwards double arrow space space dx space equals space fraction numerator 2 over denominator 1 plus straight t squared end fraction dt$
When x = 0. t = tan 0 = 0
When $straight x space equals space straight pi comma space space space straight t space equals space tan straight pi over 2 space rightwards arrow space infinity$

Also $cosx space equals space fraction numerator 1 minus tan squared begin display style straight x over 2 end style over denominator 1 plus tan squared begin display style straight x over 2 end style end fraction space equals fraction numerator 1 minus straight t squared over denominator 1 plus straight t squared end fraction$

$therefore$ $straight I space equals space integral subscript 0 superscript infinity fraction numerator begin display style fraction numerator 2 over denominator 1 plus straight t squared end fraction end style dt over denominator 6 minus begin display style fraction numerator 1 minus straight t squared over denominator 1 plus straight t squared end fraction end style end fraction space equals integral subscript 0 superscript infinity fraction numerator 2 over denominator 6 plus 6 straight t squared minus 1 plus straight t squared end fraction dt space equals space 2 integral subscript 0 superscript infinity fraction numerator 1 over denominator 7 straight t squared plus 5 end fraction dt$

$equals space 2 over 7 integral subscript 0 superscript infinity fraction numerator 1 over denominator straight t squared plus open parentheses begin display style fraction numerator square root of 5 over denominator 7 end fraction end style close parentheses squared end fraction dt space equals space 2 over 7. fraction numerator 1 over denominator square root of begin display style 5 over 7 end style end root end fraction open square brackets tan to the power of negative 1 end exponent open parentheses fraction numerator straight t over denominator square root of begin display style 5 over 7 end style end root end fraction close parentheses close square brackets subscript 0 superscript infinity equals space fraction numerator 2 over denominator square root of 35 end fraction left square bracket tan to the power of negative 1 end exponent infinity space minus space tan to the power of negative 1 end exponent 0 right square bracket space equals space fraction numerator 2 over denominator square root of 35 end fraction open square brackets straight pi over 2 minus 0 close square brackets space equals space fraction numerator straight pi over denominator square root of 35 end fraction$

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