Evaluate from Mathematics Integrals

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.
Advertisement

 Multiple Choice QuestionsLong Answer Type

111.

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

 Multiple Choice QuestionsShort Answer Type

113.

Prove that:  

209 Views

 Multiple Choice QuestionsLong 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

Advertisement
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

Advertisement

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


Let I = 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
Put   1 space equals straight r space cos space straight alpha                                          ...(1)
and   square root of 3 space equals space straight r space sin space straight alpha                                      ...(2)
Squaring and adding (1) and (2), we get
                       4 space equals space straight r squared                       rightwards double arrow space space space straight r space equals space 2
Dividing (2) by (1), tan space straight alpha space equals space square root of 3             rightwards double arrow space space space straight a space equals space straight pi over 3
therefore       straight I space equals space 1 over straight r integral subscript 0 superscript straight pi over 2 end superscript fraction numerator 1 over denominator sinx space cosα space plus space cosx space sinα end fraction dx space equals space 1 half integral subscript 0 superscript straight pi over 2 end superscript fraction numerator 1 over denominator sin left parenthesis straight x plus straight alpha right parenthesis end fraction dx
              equals space 1 half integral subscript 0 superscript straight pi over 2 end superscript cosec left parenthesis straight x plus straight alpha right parenthesis space dx space equals space 1 half open square brackets negative log space open vertical bar cosec left parenthesis straight x plus straight alpha right parenthesis plus cot left parenthesis straight x plus straight a right parenthesis close vertical bar close square brackets subscript 0 superscript straight pi over 2 end superscript

             equals space minus 1 half open square brackets log space open vertical bar cosec space open parentheses straight pi over 2 plus straight pi over 3 close parentheses plus cot space open parentheses straight pi over 2 plus straight pi over 3 close parentheses close vertical bar minus log open vertical bar cosec straight pi over 3 plus cot straight pi over 3 close vertical bar close square brackets

     equals negative 1 half open square brackets log space open vertical bar sec straight pi over 3 minus tan straight pi over 3 close vertical bar minus log open vertical bar cosec straight pi over 3 plus cot straight pi over 3 close vertical bar close square brackets

      equals negative 1 half open square brackets log space open vertical bar 2 minus square root of 3 close vertical bar minus log space open vertical bar fraction numerator 2 over denominator square root of 3 end fraction plus fraction numerator 1 over denominator square root of 3 end fraction close vertical bar close square brackets
equals negative 1 half open square brackets log left parenthesis 2 minus square root of 3 right parenthesis space minus space log square root of 3 close square brackets
equals space minus 1 half log open parentheses fraction numerator 2 minus square root of 3 over denominator square root of 3 end fraction close parentheses

120 Views

Advertisement

 Multiple Choice QuestionsShort 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

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

Advertisement
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


128 Views

Advertisement