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

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Question 3 (Supplementary NCERT) ∫1▒𝑥 √(1+𝑥−𝑥^2 ) 𝑑𝑥 ∫1▒𝑥 √(1+𝑥−𝑥^2 ) 𝑑𝑥 We can write it as:- x = A [𝑑/𝑑𝑥 (1+𝑥−𝑥^2 )]+ B x = A [0+1−2𝑥]+ B x = A [1−2𝑥]+ B x = "A"−2"A" 𝑥+ B x = −2"A" 𝑥+(𝐴+𝐵) Comparing x and constant term Thus, we can write x = A [1−2𝑥] + B x = ((−1)/2)[1−2𝑥] + 1 x = (−2A) x 𝑥/𝑥 = −2A 1 = −2A A = (−1)/2 0 = A + B B = −A B = −((−1)/2) B = 1/2 Integrating ∫1▒〖𝑥√(1+𝑥−𝑥^2 ) 〗 𝑑𝑥 = ∫1▒〖[(−1/2)[1−2𝑥]+1/2] 〗 √(1+𝑥−𝑥^2 ) 𝑑𝑥 = ∫1▒〖[(−1/2)[1 −2𝑥] √(1+𝑥−𝑥^2 )+1/2 √(1+𝑥−𝑥^2 )] 〗 𝑑𝑥 = ∫1▒〖(−1/2)[1−2𝑥] √(1+𝑥−𝑥^2 ) 𝑑𝑥+〗 ∫1▒〖1/2 √(1−𝑥−𝑥^2 )〗 𝑑𝑥 = −1/2 ∫1▒〖[1−2𝑥] √(1+𝑥−𝑥^2 ) 𝑑𝑥+〗 1/2 ∫1▒√(1+𝑥−𝑥^2 ) 𝑑𝑥 Solving 𝑰_𝟏 I_1 = (−1)/2 ∫1▒〖[1−2𝑥] √(1+𝑥−𝑥^2 )〗 𝑑𝑥 Let 1 + 𝑥 − 𝑥^2 = t Diff. both sides w.r.t.x 0 + 1 −2x = 𝑑𝑡/𝑑𝑥 (1 − 2x) dx = dt dx = 𝑑𝑡/(1 − 2𝑥) Thus, our equation becomes I_1 = (−1)/2 ∫1▒〖[1−2𝑥] √(1+𝑥−𝑥^2 )〗 𝑑𝑥 Putting the value if (1+𝑥−𝑥^2) and dx, we get I_1 = (−1)/2 ∫1▒〖[1−2𝑥] √𝑡〗. 𝑑𝑥 I_1 = (−1)/2 ∫1▒〖[1−2𝑥] √𝑡〗. 𝑑𝑡/[1 − 2𝑥] 𝐼_1 = (−1)/2 ∫1▒√𝑡. 𝑑𝑡 I_1 = (−1)/2 ∫1▒〖(𝑡)〗^(1/2) 𝑑𝑡 I_1 = (−1)/2 〖𝑡 〗^(1/2 + 1)/((1/2 + 1) )+ C_1 I_1 = (−1)/2 〖𝑡 〗^(3/2 )/((3/2) )+ C_1 I_1 = (−1)/3 〖𝑡 〗^(3/2 )+ C_1 I_1 = (−1)/3 〖(1−𝑥 −𝑥^2 ) 〗^(3/2 )+ C_1 ("Using t = " 1−𝑥 −𝑥^2 ) Solving 𝑰_𝟐 I_2 = 1/2 ∫1▒√(1+𝑥−𝑥^2 ) 𝑑𝑥 I_2 = 1/2 ∫1▒√(−(𝑥^2−𝑥−1)) 𝑑𝑥 I_2 = 1/2 ∫1▒√(−[𝑥^2−2(𝑥)(1/2)−1] ) 𝑑𝑥 I_2 = 1/2 ∫1▒〖√(−[𝑥^2−2(𝑥)(1/2)+(1/2)^2−(1/2)^2−1] ) 〗 𝑑𝑥 I_2 = 1/2 ∫1▒〖√(−[(𝑥−1/2)^2−(1/2)^2−1] ) 〗 𝑑𝑥 I_2 = 1/2 ∫1▒〖√(−[〖 (𝑥−1/2)〗^2+(−1 −4)/4] ) 〗 𝑑𝑥 I_2 = 1/2 ∫1▒〖√(−[〖 (𝑥−1/2)〗^2 −5/4] ) 〗 𝑑𝑥 I_2 = 1/2 ∫1▒〖√(5/4 〖 −(𝑥−1/2)〗^2 ) 〗 𝑑𝑥 I_2 = 1/2 ∫1▒〖√((√5/4)^2 〖 −(𝑥−1/2)〗^2 ) 〗 𝑑𝑥 I_2 = 1/2 ((𝑥 − 1/2)/2 √((√5/4)^2 〖 −(𝑥−1/2)〗^2 )+(√5/2)^2/2 〖𝑠𝑖𝑛〗^(−1) ((𝑥 + 1/2)/(√5/2))+ C_2 ) I_2 = 1/2 (((2𝑥 −1)/2)/2 √(5/4−[𝑥^2 +1/4−2𝑥(1/2)] ) +(5/4)/2 〖𝑠𝑖𝑛〗^(−1) (((2𝑥 − 1)/2)/(√5/2))+ C_2 ) It is of form √(𝑎^2−𝑥^2 ) 𝑑𝑥=1/2 𝑥√(𝑎^2−𝑥^2 )+𝑎^2/2 〖𝑠𝑖𝑛〗^(−1) 𝑥/𝑎+ C_2 Replacing x by (x – 1/2) and a by √5/2 , we get I_2 = 1/2 ((2𝑥 −1)/4 √(5/4−[𝑥^2 +1/4 −𝑥] ) +5/8 〖𝑠𝑖𝑛〗^(−1) ((2𝑥 − 1)/√5)+ C_2 ) I_2 " = " (2𝑥 −1)/8 √(1+𝑥+𝑥^2 ) +5/16 〖𝑠𝑖𝑛〗^(−1) ((2𝑥 − 1)/√5)+C_3 Putting the value of I_1 and I_2 in (1) ∫1▒𝑥 √(1+ 𝑥−𝑥^2 ) d𝑥 = (−1)/2 ∫1▒〖[1−2𝑥] √(1+𝑥−𝑥^2 )〗 𝑑𝑥+∫1▒√(1+𝑥−𝑥^2 ) 𝑑𝑥 = (−1)/3 〖(1+𝑥−𝑥^2)〗^(3/2) + C_1 + ((2𝑥 −1))/8 √(1+𝑥−𝑥^2 )+5/16 〖𝑠𝑖𝑛〗^(−1) ((2𝑥 − 1)/√5)+ C_3 = (−𝟏)/𝟑 〖(𝟏+𝒙−𝒙^𝟐)〗^(𝟑/𝟐) +𝟏/𝟖 (𝟐𝒙 −𝟏) √(𝟏+𝒙−𝒙^𝟐 )+ 𝟓/𝟏𝟔 〖𝒔𝒊𝒏〗^(−𝟏) ((𝟐𝒙 − 𝟏)/√𝟓)+ 𝑪

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.