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

  1. Chapter 7 Class 12 Integrals (Term 2)
  2. Serial order wise

Transcript

Example 25 (Supplementary NCERT) ∫1β–’π‘₯ √(1+π‘₯βˆ’π‘₯^2 ) 𝑑π‘₯ ∫1β–’π‘₯ √(1+π‘₯βˆ’π‘₯^2 ) 𝑑π‘₯ We can write it as:- x = A [𝑑/𝑑π‘₯ (1+π‘₯βˆ’π‘₯^2 )]+ B x = A [0+1βˆ’2π‘₯]+ B x = A [1βˆ’2π‘₯]+ B x = "A"βˆ’2"A" π‘₯+ B x = βˆ’2"A" π‘₯+(𝐴+𝐡) Comparing x and constant term Thus, we can write x = A [1βˆ’2π‘₯] + B x = ((βˆ’1)/2)[1βˆ’2π‘₯] + 1 x = (βˆ’2A) x π‘₯/π‘₯ = βˆ’2A 1 = βˆ’2A A = (βˆ’1)/2 0 = A + B B = βˆ’A B = βˆ’((βˆ’1)/2) B = 1/2 Integrating ∫1β–’γ€–π‘₯√(1+π‘₯βˆ’π‘₯^2 ) γ€— 𝑑π‘₯ = ∫1β–’γ€–[(βˆ’1/2)[1βˆ’2π‘₯]+1/2] γ€— √(1+π‘₯βˆ’π‘₯^2 ) 𝑑π‘₯ = ∫1β–’γ€–[(βˆ’1/2)[1 βˆ’2π‘₯] √(1+π‘₯βˆ’π‘₯^2 )+1/2 √(1+π‘₯βˆ’π‘₯^2 )] γ€— 𝑑π‘₯ = ∫1β–’γ€–(βˆ’1/2)[1βˆ’2π‘₯] √(1+π‘₯βˆ’π‘₯^2 ) 𝑑π‘₯+γ€— ∫1β–’γ€–1/2 √(1βˆ’π‘₯βˆ’π‘₯^2 )γ€— 𝑑π‘₯ = βˆ’1/2 ∫1β–’γ€–[1βˆ’2π‘₯] √(1+π‘₯βˆ’π‘₯^2 ) 𝑑π‘₯+γ€— 1/2 ∫1β–’βˆš(1+π‘₯βˆ’π‘₯^2 ) 𝑑π‘₯ Solving 𝑰_𝟏 I_1 = (βˆ’1)/2 ∫1β–’γ€–[1βˆ’2π‘₯] √(1+π‘₯βˆ’π‘₯^2 )γ€— 𝑑π‘₯ Let 1 + π‘₯ βˆ’ π‘₯^2 = t Diff. both sides w.r.t.x 0 + 1 βˆ’2x = 𝑑𝑑/𝑑π‘₯ (1 βˆ’ 2x) dx = dt dx = 𝑑𝑑/(1 βˆ’ 2π‘₯) Thus, our equation becomes I_1 = (βˆ’1)/2 ∫1β–’γ€–[1βˆ’2π‘₯] √(1+π‘₯βˆ’π‘₯^2 )γ€— 𝑑π‘₯ Putting the value if (1+π‘₯βˆ’π‘₯^2) and dx, we get I_1 = (βˆ’1)/2 ∫1β–’γ€–[1βˆ’2π‘₯] βˆšπ‘‘γ€—. 𝑑π‘₯ I_1 = (βˆ’1)/2 ∫1β–’γ€–[1βˆ’2π‘₯] βˆšπ‘‘γ€—. 𝑑𝑑/[1 βˆ’ 2π‘₯] 𝐼_1 = (βˆ’1)/2 ∫1β–’βˆšπ‘‘. 𝑑𝑑 I_1 = (βˆ’1)/2 ∫1β–’γ€–(𝑑)γ€—^(1/2) 𝑑𝑑 I_1 = (βˆ’1)/2 〖𝑑 γ€—^(1/2 + 1)/((1/2 + 1) )+ C_1 I_1 = (βˆ’1)/2 〖𝑑 γ€—^(3/2 )/((3/2) )+ C_1 I_1 = (βˆ’1)/3 〖𝑑 γ€—^(3/2 )+ C_1 I_1 = (βˆ’1)/3 γ€–(1βˆ’π‘₯ βˆ’π‘₯^2 ) γ€—^(3/2 )+ C_1 ("Using t = " 1βˆ’π‘₯ βˆ’π‘₯^2 ) Solving 𝑰_𝟐 I_2 = 1/2 ∫1β–’βˆš(1+π‘₯βˆ’π‘₯^2 ) 𝑑π‘₯ I_2 = 1/2 ∫1β–’βˆš(βˆ’(π‘₯^2βˆ’π‘₯βˆ’1)) 𝑑π‘₯ I_2 = 1/2 ∫1β–’βˆš(βˆ’[π‘₯^2βˆ’2(π‘₯)(1/2)βˆ’1] ) 𝑑π‘₯ I_2 = 1/2 ∫1β–’γ€–βˆš(βˆ’[π‘₯^2βˆ’2(π‘₯)(1/2)+(1/2)^2βˆ’(1/2)^2βˆ’1] ) γ€— 𝑑π‘₯ I_2 = 1/2 ∫1β–’γ€–βˆš(βˆ’[(π‘₯βˆ’1/2)^2βˆ’(1/2)^2βˆ’1] ) γ€— 𝑑π‘₯ I_2 = 1/2 ∫1β–’γ€–βˆš(βˆ’[γ€– (π‘₯βˆ’1/2)γ€—^2+(βˆ’1 βˆ’4)/4] ) γ€— 𝑑π‘₯ I_2 = 1/2 ∫1β–’γ€–βˆš(βˆ’[γ€– (π‘₯βˆ’1/2)γ€—^2 βˆ’5/4] ) γ€— 𝑑π‘₯ I_2 = 1/2 ∫1β–’γ€–βˆš(5/4 γ€– βˆ’(π‘₯βˆ’1/2)γ€—^2 ) γ€— 𝑑π‘₯ I_2 = 1/2 ∫1β–’γ€–βˆš((√5/4)^2 γ€– βˆ’(π‘₯βˆ’1/2)γ€—^2 ) γ€— 𝑑π‘₯ I_2 = 1/2 ((π‘₯ βˆ’ 1/2)/2 √((√5/4)^2 γ€– βˆ’(π‘₯βˆ’1/2)γ€—^2 )+(√5/2)^2/2 〖𝑠𝑖𝑛〗^(βˆ’1) ((π‘₯ + 1/2)/(√5/2))+ C_2 ) I_2 = 1/2 (((2π‘₯ βˆ’1)/2)/2 √(5/4βˆ’[π‘₯^2 +1/4βˆ’2π‘₯(1/2)] ) +(5/4)/2 〖𝑠𝑖𝑛〗^(βˆ’1) (((2π‘₯ βˆ’ 1)/2)/(√5/2))+ C_2 ) It is of form √(π‘Ž^2βˆ’π‘₯^2 ) 𝑑π‘₯=1/2 π‘₯√(π‘Ž^2βˆ’π‘₯^2 )+π‘Ž^2/2 〖𝑠𝑖𝑛〗^(βˆ’1) π‘₯/π‘Ž+ C_2 Replacing x by (x – 1/2) and a by √5/2 , we get I_2 = 1/2 ((2π‘₯ βˆ’1)/4 √(5/4βˆ’[π‘₯^2 +1/4 βˆ’π‘₯] ) +5/8 〖𝑠𝑖𝑛〗^(βˆ’1) ((2π‘₯ βˆ’ 1)/√5)+ C_2 ) I_2 " = " (2π‘₯ βˆ’1)/8 √(1+π‘₯+π‘₯^2 ) +5/16 〖𝑠𝑖𝑛〗^(βˆ’1) ((2π‘₯ βˆ’ 1)/√5)+C_3 Putting the value of I_1 and I_2 in (1) ∫1β–’π‘₯ √(1+ π‘₯βˆ’π‘₯^2 ) dπ‘₯ = (βˆ’1)/2 ∫1β–’γ€–[1βˆ’2π‘₯] √(1+π‘₯βˆ’π‘₯^2 )γ€— 𝑑π‘₯+∫1β–’βˆš(1+π‘₯βˆ’π‘₯^2 ) 𝑑π‘₯ = (βˆ’1)/3 γ€–(1+π‘₯βˆ’π‘₯^2)γ€—^(3/2) + C_1 + ((2π‘₯ βˆ’1))/8 √(1+π‘₯βˆ’π‘₯^2 )+5/16 〖𝑠𝑖𝑛〗^(βˆ’1) ((2π‘₯ βˆ’ 1)/√5)+ C_3 = (βˆ’πŸ)/πŸ‘ γ€–(𝟏+π’™βˆ’π’™^𝟐)γ€—^(πŸ‘/𝟐) +𝟏/πŸ– (πŸπ’™ βˆ’πŸ) √(𝟏+π’™βˆ’π’™^𝟐 )+ πŸ“/πŸπŸ” γ€–π’”π’Šπ’γ€—^(βˆ’πŸ) ((πŸπ’™ βˆ’ 𝟏)/βˆšπŸ“)+ π‘ͺ

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.