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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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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.