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Evaluate: $integral subscript 0 superscript straight pi over 4 end superscript space secx square root of fraction numerator 1 minus sin space straight x over denominator 1 plus sin space straight x end fraction end root dx.$

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112.

Evaluate: $integral subscript 0 superscript straight a space sin to the power of negative 1 end exponent space open parentheses square root of fraction numerator straight x over denominator straight a plus straight x end fraction end root close parentheses dx.$

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Short Answer Type

113.

Prove that:

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Long Answer Type

114.

Evaluate:
$integral subscript 0 superscript straight pi over 2 end superscript fraction numerator dx over denominator 4 cosx plus 2 sinx end fraction.$

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115.

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

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116.

Evaluate
$integral subscript 0 superscript straight pi over 2 end superscript fraction numerator 1 over denominator sinx plus square root of 3 cosx end fraction dx$

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Short Answer Type

117.

Prove that:
$integral subscript 0 superscript straight pi over 2 end superscript fraction numerator dx over denominator 9 plus 16 space cos squared straight x end fraction space equals space straight pi over 30$

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118.

Prove that:
$integral subscript 0 superscript straight pi over 4 end superscript fraction numerator sin space 2 straight theta space dθ over denominator sin to the power of 4 space straight theta space plus space cos to the power of 4 straight theta end fraction space equals space straight pi over 4$

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119.

Prove that:
$integral subscript 0 superscript straight pi over 2 end superscript fraction numerator sin space 2 straight ϕ space dϕ over denominator sin to the power of 4 straight ϕ plus cos to the power of 4 straight ϕ end fraction space equals space straight pi over 2$

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120.
Prove that:
$integral subscript 0 superscript straight pi over 2 end superscript fraction numerator cosx space dx over denominator left parenthesis 1 plus sin space straight x right parenthesis space left parenthesis 2 plus sinx right parenthesis end fraction space equals space log space 4 over 3$

Let
$straight I space equals space integral subscript 0 superscript straight pi over 2 end superscript fraction numerator cosx space dx over denominator left parenthesis 1 plus sinx right parenthesis space left parenthesis 2 plus sinx right parenthesis end fraction$
Put sin x = t, ∴ cos x dx = dt
When x = 0, t = sin 0 = 0
When $straight x space equals space straight pi over 2 comma space space space straight t space equals space sin straight pi over 2 space equals space 1$
I = $integral subscript 0 superscript 1 fraction numerator dt over denominator left parenthesis 1 plus straight t right parenthesis space left parenthesis 2 plus straight t right parenthesis end fraction$ [Do not forget to change limits of integration]
$equals space integral subscript 0 superscript 1 open square brackets fraction numerator 1 over denominator left parenthesis 1 plus straight t right parenthesis space left parenthesis 2 minus 1 right parenthesis end fraction plus fraction numerator 1 over denominator left parenthesis 1 minus 2 right parenthesis thin space left parenthesis 2 plus straight t right parenthesis end fraction close square brackets dt$
$equals space integral subscript 0 superscript 1 open square brackets fraction numerator 1 over denominator 1 plus straight t end fraction minus fraction numerator 1 over denominator 2 plus straight t end fraction close square brackets dt space equals space open square brackets log space left parenthesis 1 plus straight t right parenthesis space minus space log left parenthesis 2 plus straight t right parenthesis close square brackets subscript 0 superscript 1$
$equals space open square brackets log space fraction numerator 1 plus straight t over denominator 2 plus straight t end fraction close square brackets subscript 0 superscript 1 space equals space log 2 over 3 minus log 1 half space equals space log open parentheses fraction numerator begin display style 2 over 3 end style over denominator begin display style 1 half end style end fraction close parentheses space equals space log 4 over 3$

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120. Prove that: