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

161.

Show that:
$integral subscript 0 superscript straight pi over 2 end superscript fraction numerator cos space straight x over denominator sin space straight x space plus space cos space straight x end fraction dx$

81 Views

Long Answer Type

162.

Evaluate $integral subscript straight pi over 6 end subscript superscript straight pi over 3 end superscript fraction numerator dx over denominator 1 plus square root of tanx end fraction$

92 Views

163.

Evaluate $integral subscript straight pi over 6 end subscript superscript straight pi over 3 end superscript fraction numerator square root of sin space straight x end root over denominator square root of sin space straight x end root plus square root of cos space straight x end root end fraction dx$

97 Views

Short Answer Type

164.

Evaluate $integral subscript 0 superscript straight pi space sin squared straight x space cos cubed straight x space dx.$

115 Views

165.

Evaluate $integral subscript negative straight pi over 4 end subscript superscript straight pi over 4 end superscript space sin squared straight x space dx$

105 Views

166.

Evaluate $integral subscript negative straight pi over 4 end subscript superscript straight pi over 4 end superscript cos squared xdx$

116 Views

167.

By using the properties of definite integrals, evaluate
$integral subscript negative straight pi over 2 end subscript superscript straight pi over 2 end superscript cosx space dx$

98 Views

168.
Evaluate $integral subscript negative 1 end subscript superscript 1 sin to the power of 5 straight x space cos to the power of 4 straight x space dx$

Comparing $integral subscript negative 1 end subscript superscript 1 space sin to the power of 5 straight x space cos to the power of 4 straight x space dx space with space integral subscript negative 1 end subscript superscript 1 space straight f left parenthesis straight x right parenthesis space dx comma space space we space get$
f (x) = sin⁵ x cos⁴ x
Now f (–x) = sin⁵ (–x) cos⁴ (–x) = – sin⁵x cos⁴ x = – f (x)
∴ f (x) is an odd function
$therefore space space space space straight I space equals space 0$ [ $integral subscript negative straight a end subscript superscript straight a straight f left parenthesis straight x right parenthesis space dx space equals space 0 space space if space straight f left parenthesis straight x right parenthesis$ is an odd function]
$therefore space space space integral subscript negative 1 end subscript superscript 1 space sin to the power of 5 straight x space cos to the power of 4 straight x space dx space equals space 0$

108 Views

169.

Evaluate $integral subscript negative 1 end subscript superscript 1 space straight x to the power of 17 space cos to the power of 4 straight x space dx.$

85 Views

170.

Evaluate $integral subscript negative 1 end subscript superscript 1 space log space open parentheses fraction numerator 2 plus straight x over denominator 2 minus straight x end fraction close parentheses dx.$