Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

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

93 Views

112.

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

109 Views

Short Answer Type

113.

Prove that:

209 Views

Long Answer Type

114.

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

99 Views

115.

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

122 Views

116.

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

120 Views

Short Answer Type

117.
Prove that:
$integral subscript 0 superscript straight pi over 2 end superscript fraction numerator dx over denominator 9 plus 16 space cos squared straight x end fraction space equals space straight pi over 30$

Let I = $integral subscript 0 superscript straight pi over 2 end superscript fraction numerator dx over denominator 9 plus 16 space cos squared straight x end fraction space equals space integral subscript 0 superscript straight pi over 2 end superscript fraction numerator dx over denominator 9 space left parenthesis sin squared straight x plus cos squared straight x right parenthesis plus 16 cos squared straight x end fraction$

$equals space integral subscript 0 superscript straight pi over 2 end superscript fraction numerator dx over denominator 9 space sin squared straight x plus 25 space cos squared straight x end fraction space equals space integral subscript 0 superscript straight pi over 2 end superscript fraction numerator begin display style fraction numerator 1 over denominator cos squared space straight x end fraction end style dx over denominator begin display style fraction numerator 9 space sin squared straight x over denominator cos squared straight x end fraction end style plus begin display style fraction numerator 25 space cos squared straight x over denominator cos squared straight x end fraction end style end fraction$
   $equals space integral subscript 0 superscript straight pi over 2 end superscript fraction numerator sec squared xdx over denominator 9 space tan squared straight x plus 25 end fraction$
Put tan x = t,    ∴ sec² x dx = dt
When x = 0, t = tan 0 = 0
When    $straight x space equals space straight pi over 2 comma space space straight t space equals space tan straight pi over 2 space equals space infinity$

$therefore$    $straight I space equals space integral subscript 0 superscript infinity fraction numerator dt over denominator 9 straight t squared plus 25 end fraction space equals space 1 over 9 integral subscript 0 superscript infinity fraction numerator 1 over denominator straight t squared plus begin display style 25 over 9 end style end fraction dt space equals space 1 over 9 integral subscript 0 superscript infinity fraction numerator 1 over denominator straight t squared plus open parentheses begin display style 5 over 3 end style close parentheses squared end fraction dt$

$equals space 1 over 9. fraction numerator 1 over denominator begin display style 5 over 3 end style end fraction open square brackets tan to the power of negative 1 end exponent fraction numerator straight t over denominator open parentheses begin display style 5 over 3 end style close parentheses end fraction close square brackets subscript 0 superscript infinity space equals 1 over 15 open square brackets tan to the power of negative 1 end exponent open parentheses fraction numerator 3 straight t over denominator 5 end fraction close parentheses close square brackets subscript 0 superscript infinity$

$equals space 1 over 15 open square brackets tan to the power of negative 1 end exponent infinity space minus space tan to the power of negative 1 end exponent 0 close square brackets space equals space 1 over 15 open square brackets straight pi over 2 minus 0 close square brackets space equals space straight pi over 30$

106 Views

118.

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

109 Views

119.

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

113 Views

120.

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