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

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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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