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.

Integrals

Short Answer Type

191.

Show that:
$integral subscript straight pi over 4 end subscript superscript straight pi over 4 end superscript open vertical bar sin space straight x close vertical bar dx space equals space 2 minus square root of 2$

120 Views

Long Answer Type

192.

Show that:
$integral subscript negative 1 end subscript superscript 3 over 2 end superscript open vertical bar straight x space sinπ space straight x close vertical bar space dx space equals space 3 over straight pi plus 1 over straight pi squared$

152 Views

Short Answer Type

193.

Show that:
$integral subscript 0 superscript 2 straight x square root of 2 minus straight x end root space equals fraction numerator 16 square root of 2 over denominator 15 end fraction$

147 Views

194.

By using the properties of definite integrals, evaluate the following integral:
$integral subscript 0 superscript 1 space straight x space left parenthesis 1 minus straight x right parenthesis to the power of straight n space dx$

108 Views

195.
By using the properties of definite integrals, evaluate the following integral:
$integral subscript 0 superscript straight pi over 2 end superscript fraction numerator sinx minus cosx over denominator 1 plus sinxcosx end fraction dx$

Let I = $integral subscript 0 superscript straight pi over 2 end superscript fraction numerator sinx minus cosx over denominator 1 plus sinx. cosx end fraction dx$ ...(1)
$therefore space space space space space space space straight I space space equals space integral subscript 0 superscript straight pi over 2 end superscript fraction numerator sin space open parentheses begin display style straight pi over 2 end style minus straight x close parentheses minus cos open parentheses begin display style straight pi over 2 end style minus straight x close parentheses over denominator 1 plus sin open parentheses begin display style straight pi over 2 end style minus straight x close parentheses space cos open parentheses begin display style straight pi over 2 end style minus straight x close parentheses end fraction$ $open square brackets because space integral subscript 0 superscript straight a straight f left parenthesis straight x right parenthesis space dx space equals integral subscript 0 superscript straight a straight f left parenthesis straight a minus straight x right parenthesis space dx close square brackets$

rightwards double arrow space space straight I space equals space integral subscript 0 superscript straight pi over 2 end superscript fraction numerator cosx minus space sinx over denominator 1 plus cosx space sinx end fraction dx space space space rightwards double arrow space space straight I space equals space minus integral subscript 0 superscript straight pi over 2 end superscript fraction numerator sinx minus cosx over denominator 1 plus sinx space cosx end fraction dx rightwards double arrow space space straight I space equals space minus straight I space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because space of space left parenthesis 1 right parenthesis close square brackets rightwards double arrow space 2 space straight I space equals space 0 space rightwards double arrow space straight I space equals space 0

134 Views

196.

Show that:
$integral subscript 0 superscript straight pi over 2 end superscript 2 straight x space log space tanx space dx space equals space 0$

110 Views

197.

By using the properties of definite integrals, evaluate the following integral:
$integral subscript 0 superscript straight pi over 2 end superscript left parenthesis 2 space log space sinx space minus space log space sin space 2 straight x right parenthesis space dx$

410 Views

198.

Show that:
$integral subscript 0 superscript straight pi fraction numerator straight x space tanx over denominator secx space plus space cosx end fraction dx space equals space straight pi squared over 4$

100 Views

199.

Show that:
$integral subscript 0 superscript infinity log space open parentheses straight x plus 1 over straight x close parentheses. space fraction numerator dx over denominator 1 plus straight x squared end fraction space equals space straight pi space log space 2$

132 Views

200.

Show that:
$integral subscript 0 superscript 1 fraction numerator log space left parenthesis 1 plus straight x right parenthesis over denominator 1 plus straight x squared end fraction space equals space straight pi over 8 space log space 2$