Question 3 (Supplementary NCERT) - Examples - Chapter 7 Class 12 Integrals

Last updated at December 16, 2024 by

Example 25 (Supplementary NCERT) - Integrate x root(1 + x - x2) dx

Example 25 (Supplementary NCERT) - Chapter 7 Class 12 Integrals - Part 2
Example 25 (Supplementary NCERT) - Chapter 7 Class 12 Integrals - Part 3
Example 25 (Supplementary NCERT) - Chapter 7 Class 12 Integrals - Part 4
Example 25 (Supplementary NCERT) - Chapter 7 Class 12 Integrals - Part 5 Example 25 (Supplementary NCERT) - Chapter 7 Class 12 Integrals - Part 6 Example 25 (Supplementary NCERT) - Chapter 7 Class 12 Integrals - Part 7 Example 25 (Supplementary NCERT) - Chapter 7 Class 12 Integrals - Part 8 Example 25 (Supplementary NCERT) - Chapter 7 Class 12 Integrals - Part 9

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Question 3 (Supplementary NCERT) āˆ«1ā–’š‘„ āˆš(1+š‘„āˆ’š‘„^2 ) š‘‘š‘„ āˆ«1ā–’š‘„ āˆš(1+š‘„āˆ’š‘„^2 ) š‘‘š‘„ We can write it as:- x = A [š‘‘/š‘‘š‘„ (1+š‘„āˆ’š‘„^2 )]+ B x = A [0+1āˆ’2š‘„]+ B x = A [1āˆ’2š‘„]+ B x = "A"āˆ’2"A" š‘„+ B x = āˆ’2"A" š‘„+(š“+šµ) Comparing x and constant term Thus, we can write x = A [1āˆ’2š‘„] + B x = ((āˆ’1)/2)[1āˆ’2š‘„] + 1 x = (āˆ’2A) x š‘„/š‘„ = āˆ’2A 1 = āˆ’2A A = (āˆ’1)/2 0 = A + B B = āˆ’A B = āˆ’((āˆ’1)/2) B = 1/2 Integrating āˆ«1ā–’怖š‘„āˆš(1+š‘„āˆ’š‘„^2 ) 怗 š‘‘š‘„ = āˆ«1ā–’怖[(āˆ’1/2)[1āˆ’2š‘„]+1/2] 怗 āˆš(1+š‘„āˆ’š‘„^2 ) š‘‘š‘„ = āˆ«1ā–’怖[(āˆ’1/2)[1 āˆ’2š‘„] āˆš(1+š‘„āˆ’š‘„^2 )+1/2 āˆš(1+š‘„āˆ’š‘„^2 )] 怗 š‘‘š‘„ = āˆ«1ā–’怖(āˆ’1/2)[1āˆ’2š‘„] āˆš(1+š‘„āˆ’š‘„^2 ) š‘‘š‘„+怗 āˆ«1ā–’怖1/2 āˆš(1āˆ’š‘„āˆ’š‘„^2 )怗 š‘‘š‘„ = āˆ’1/2 āˆ«1ā–’怖[1āˆ’2š‘„] āˆš(1+š‘„āˆ’š‘„^2 ) š‘‘š‘„+怗 1/2 āˆ«1ā–’āˆš(1+š‘„āˆ’š‘„^2 ) š‘‘š‘„ Solving š‘°_šŸ I_1 = (āˆ’1)/2 āˆ«1ā–’怖[1āˆ’2š‘„] āˆš(1+š‘„āˆ’š‘„^2 )怗 š‘‘š‘„ Let 1 + š‘„ āˆ’ š‘„^2 = t Diff. both sides w.r.t.x 0 + 1 āˆ’2x = š‘‘š‘”/š‘‘š‘„ (1 āˆ’ 2x) dx = dt dx = š‘‘š‘”/(1 āˆ’ 2š‘„) Thus, our equation becomes I_1 = (āˆ’1)/2 āˆ«1ā–’怖[1āˆ’2š‘„] āˆš(1+š‘„āˆ’š‘„^2 )怗 š‘‘š‘„ Putting the value if (1+š‘„āˆ’š‘„^2) and dx, we get I_1 = (āˆ’1)/2 āˆ«1ā–’怖[1āˆ’2š‘„] āˆšš‘”怗. š‘‘š‘„ I_1 = (āˆ’1)/2 āˆ«1ā–’怖[1āˆ’2š‘„] āˆšš‘”怗. š‘‘š‘”/[1 āˆ’ 2š‘„] š¼_1 = (āˆ’1)/2 āˆ«1ā–’āˆšš‘”. š‘‘š‘” I_1 = (āˆ’1)/2 āˆ«1ā–’怖(š‘”)怗^(1/2) š‘‘š‘” I_1 = (āˆ’1)/2 怖š‘” 怗^(1/2 + 1)/((1/2 + 1) )+ C_1 I_1 = (āˆ’1)/2 怖š‘” 怗^(3/2 )/((3/2) )+ C_1 I_1 = (āˆ’1)/3 怖š‘” 怗^(3/2 )+ C_1 I_1 = (āˆ’1)/3 怖(1āˆ’š‘„ āˆ’š‘„^2 ) 怗^(3/2 )+ C_1 ("Using t = " 1āˆ’š‘„ āˆ’š‘„^2 ) Solving š‘°_šŸ I_2 = 1/2 āˆ«1ā–’āˆš(1+š‘„āˆ’š‘„^2 ) š‘‘š‘„ I_2 = 1/2 āˆ«1ā–’āˆš(āˆ’(š‘„^2āˆ’š‘„āˆ’1)) š‘‘š‘„ I_2 = 1/2 āˆ«1ā–’āˆš(āˆ’[š‘„^2āˆ’2(š‘„)(1/2)āˆ’1] ) š‘‘š‘„ I_2 = 1/2 āˆ«1ā–’怖āˆš(āˆ’[š‘„^2āˆ’2(š‘„)(1/2)+(1/2)^2āˆ’(1/2)^2āˆ’1] ) 怗 š‘‘š‘„ I_2 = 1/2 āˆ«1ā–’怖āˆš(āˆ’[(š‘„āˆ’1/2)^2āˆ’(1/2)^2āˆ’1] ) 怗 š‘‘š‘„ I_2 = 1/2 āˆ«1ā–’怖āˆš(āˆ’[怖 (š‘„āˆ’1/2)怗^2+(āˆ’1 āˆ’4)/4] ) 怗 š‘‘š‘„ I_2 = 1/2 āˆ«1ā–’怖āˆš(āˆ’[怖 (š‘„āˆ’1/2)怗^2 āˆ’5/4] ) 怗 š‘‘š‘„ I_2 = 1/2 āˆ«1ā–’怖āˆš(5/4 怖 āˆ’(š‘„āˆ’1/2)怗^2 ) 怗 š‘‘š‘„ I_2 = 1/2 āˆ«1ā–’怖āˆš((āˆš5/4)^2 怖 āˆ’(š‘„āˆ’1/2)怗^2 ) 怗 š‘‘š‘„ I_2 = 1/2 ((š‘„ āˆ’ 1/2)/2 āˆš((āˆš5/4)^2 怖 āˆ’(š‘„āˆ’1/2)怗^2 )+(āˆš5/2)^2/2 怖š‘ š‘–š‘›ć€—^(āˆ’1) ((š‘„ + 1/2)/(āˆš5/2))+ C_2 ) I_2 = 1/2 (((2š‘„ āˆ’1)/2)/2 āˆš(5/4āˆ’[š‘„^2 +1/4āˆ’2š‘„(1/2)] ) +(5/4)/2 怖š‘ š‘–š‘›ć€—^(āˆ’1) (((2š‘„ āˆ’ 1)/2)/(āˆš5/2))+ C_2 ) It is of form āˆš(š‘Ž^2āˆ’š‘„^2 ) š‘‘š‘„=1/2 š‘„āˆš(š‘Ž^2āˆ’š‘„^2 )+š‘Ž^2/2 怖š‘ š‘–š‘›ć€—^(āˆ’1) š‘„/š‘Ž+ C_2 Replacing x by (x ā€“ 1/2) and a by āˆš5/2 , we get I_2 = 1/2 ((2š‘„ āˆ’1)/4 āˆš(5/4āˆ’[š‘„^2 +1/4 āˆ’š‘„] ) +5/8 怖š‘ š‘–š‘›ć€—^(āˆ’1) ((2š‘„ āˆ’ 1)/āˆš5)+ C_2 ) I_2 " = " (2š‘„ āˆ’1)/8 āˆš(1+š‘„+š‘„^2 ) +5/16 怖š‘ š‘–š‘›ć€—^(āˆ’1) ((2š‘„ āˆ’ 1)/āˆš5)+C_3 Putting the value of I_1 and I_2 in (1) āˆ«1ā–’š‘„ āˆš(1+ š‘„āˆ’š‘„^2 ) dš‘„ = (āˆ’1)/2 āˆ«1ā–’怖[1āˆ’2š‘„] āˆš(1+š‘„āˆ’š‘„^2 )怗 š‘‘š‘„+āˆ«1ā–’āˆš(1+š‘„āˆ’š‘„^2 ) š‘‘š‘„ = (āˆ’1)/3 怖(1+š‘„āˆ’š‘„^2)怗^(3/2) + C_1 + ((2š‘„ āˆ’1))/8 āˆš(1+š‘„āˆ’š‘„^2 )+5/16 怖š‘ š‘–š‘›ć€—^(āˆ’1) ((2š‘„ āˆ’ 1)/āˆš5)+ C_3 = (āˆ’šŸ)/šŸ‘ 怖(šŸ+š’™āˆ’š’™^šŸ)怗^(šŸ‘/šŸ) +šŸ/šŸ– (šŸš’™ āˆ’šŸ) āˆš(šŸ+š’™āˆ’š’™^šŸ )+ šŸ“/šŸšŸ” 怖š’”š’Šš’ć€—^(āˆ’šŸ) ((šŸš’™ āˆ’ šŸ)/āˆššŸ“)+ š‘Ŗ

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo