Question 1 - Chapter 7 Class 12 Integrals (Important Question)
Last updated at April 16, 2024 by Teachoo
Chapter 7 Class 12 Integrals
Ex 7.1, 18 Important
Ex 7.1, 20
Ex 7.2, 20 Important
Ex 7.2, 26 Important
Ex 7.2, 35
Ex 7.2, 36 Important
Ex 7.3, 6 Important
Ex 7.3, 13 Important
Ex 7.3, 18 Important
Ex 7.3, 22 Important
Ex 7.3, 24 (MCQ) Important
Example 9 (i)
Example 10 (i)
Ex 7.4, 8 Important
Ex 7.4, 15 Important
Ex 7.4, 21 Important
Ex 7.4, 22
Ex 7.4, 25 (MCQ) Important
Example 15 Important
Ex 7.5, 9 Important
Ex 7.5, 11 Important
Ex 7.5, 17
Ex 7.5, 18 Important
Ex 7.5, 21 Important
Example 20 Important
Example 22 Important
Ex 7.6, 13 Important
Ex 7.6, 14 Important
Ex 7.6, 18 Important
Ex 7.6, 19
Ex 7.6, 24 (MCQ) Important
Ex 7.7, 5 Important
Ex 7.7, 10
Ex 7.7, 11 Important
Question 1 Important Deleted for CBSE Board 2024 Exams You are here
Question 4 Important Deleted for CBSE Board 2024 Exams
Question 6 Important Deleted for CBSE Board 2024 Exams
Example 25 (i)
Ex 7.8, 15
Ex 7.8, 16 Important
Ex 7.8, 20 Important
Ex 7.8, 22 (MCQ)
Ex 7.9, 4
Ex 7.9, 7 Important
Ex 7.9, 8
Ex 7.9, 9 (MCQ) Important
Example 28 Important
Example 32 Important
Example 34 Important
Ex 7.10,8 Important
Ex 7.10, 18 Important
Example 38 Important
Example 39 Important
Example 42 Important
Misc 18 Important
Misc 8 Important
Question 1 Important Deleted for CBSE Board 2024 Exams You are here
Misc 23 Important
Misc 29 Important
Question 2 Important Deleted for CBSE Board 2024 Exams
Misc 38 (MCQ) Important
Question 4 (MCQ) Important Deleted for CBSE Board 2024 Exams
Integration Formula Sheet - Chapter 41 Class 41 Formulas Important
Chapter 7 Class 12 Integrals
Last updated at April 16, 2024 by Teachoo
Question 1 Integrate the function (sin^(โ1)โกโ๐ฅ โ cos^(โ1)โกโ๐ฅ)/(sin^(โ1)โกโ๐ฅ + cos^(โ1)โกโ๐ฅ ) , ๐ฅโ[0, 1] Let ๐ผ = โซ1โ(sin^(โ1)โกโ๐ฅ โ cos^(โ1)โกโ๐ฅ)/(sin^(โ1)โกโ๐ฅ + cos^(โ1)โกโ๐ฅ ) ๐๐ฅ We can write as (sin^(โ1)โกโ๐ฅ โ cos^(โ1)โกโ๐ฅ)/(sin^(โ1)โกโ๐ฅ + cos^(โ1)โกโ๐ฅ ) = (sin^(โ1)โกโ๐ฅ โ (๐/2 " โ" ใ ๐ ๐๐ใ^(โ1)โกโ๐ฅ ))/(๐/2) We know that ใ๐ ๐๐ใ^(โ1)โก๐ฅ+ใ๐๐๐ ใ^(โ1)โก๐ฅ=๐/2 or ใ๐๐๐ ใ^(โ1)โก๐ฅ=๐/2 โใ ๐ ๐๐ใ^(โ1)โก๐ฅ = (2/๐)(sin^(โ1)โกโ๐ฅ โ๐/2 " +" ใ ๐ ๐๐ใ^(โ1)โกโ๐ฅ ) = 2/๐ (2 sin^(โ1)โกโ๐ฅ โ๐/2) = 2/๐ ร2 sin^(โ1)โกโ๐ฅโ 2/๐ร๐/2 = 4/๐ sin^(โ1)โกโ๐ฅโ1 Integrating ๐ค.๐.๐ก.๐ฅ โซ1โ(sin^(โ1)โกโ๐ฅ โ cos^(โ1)โกโ๐ฅ)/(sin^(โ1)โกโ๐ฅ + cos^(โ1)โกโ๐ฅ ) ๐๐ฅ=โซ1โ(4/๐ sin^(โ1)โกโ๐ฅโ1) ๐๐ฅ = โซ1โใ4/๐ sin^(โ1)โกโ๐ฅ ใ ๐๐ฅโโซ1โ๐๐ฅ = 4/๐ โซ1โsin^(โ1)โกโ๐ฅ ๐๐ฅโ๐ฅ+๐ถ1 Let ๐ผ1=โซ1โsin^(โ1)โกโ๐ฅ ๐๐ฅ Hence, I = 4/๐ ๐ผ1โ๐ฅ+๐ถ1 Solving ๐_๐ ๐ผ1 = โซ1โsin^(โ1)โกโ๐ฅ ๐๐ฅ Put โ๐ฅ=๐ก ๐ฅ=๐ก^2 Differentiating ๐ค.๐.๐ก.๐ฅ ๐๐ฅ/๐๐ฅ = (๐๐ก^2)/๐๐ฅ 1 = 2๐ก ๐๐ก/๐๐ฅ ๐๐ฅ = 2๐ก ๐๐ก Therefore โซ1โsin^(โ1)โกโ๐ฅ ๐๐ฅ=โซ1โsin^(โ1)โก๐ก .2๐ก ๐๐ก =2โซ1โsin^(โ1)โก๐ก .๐ก ๐๐ก =2โซ1โใ๐ก sin^(โ1)โกใ๐ก ใ ใ ๐๐ก =2[sin^(โ1)โกใ๐ก ใ โซ1โ๐ก ๐๐กโโซ1โ((๐/๐๐ก sin^(โ1)โก๐ก ) โซ1โใ๐ก ๐๐กใ) ๐๐ก Now we know that โซ1โใ๐(๐ฅ) ๐โก(๐ฅ) ใ ๐๐ฅ=๐(๐ฅ) โซ1โ๐(๐ฅ) ๐๐ฅโโซ1โ(๐^โฒ (๐ฅ) โซ1โ๐(๐ฅ) ๐๐ฅ) ๐๐ฅ Putting f(x) = t and g(x) = sinโ1 t =2[sin^(โ1)โกใ๐ก ใ ๐ก^2/2 โโซ1โ1/โ(1 โใ ๐กใ^2 ) ร๐ก^2/2 ๐๐ก+๐ถ] =2ร๐ก^2/2 ใ sin^(โ1)ใโก๐กโ2รโซ1โใ1/2 ร๐ก^2/โ(1 โใ ๐กใ^2 )ใ ๐๐ก+๐ถ = ๐ก^2 sin^(โ1)โก๐กโโซ1โ๐ก^2/โ(1 โใ ๐กใ^2 ) ๐๐ก+๐ถ = ๐ก^2 sin^(โ1)โก๐ก+โซ1โ(โ๐ก^2)/โ(1 โใ ๐กใ^2 ) ๐๐ก+๐ถ Solving โซ1โใโ ๐ใ^๐/โ(๐ โใ ๐ใ^๐ ) ๐ ๐ We can write (โ ๐ก^2)/โ(1 โใ ๐กใ^2 ) =(ใโ ๐กใ^2 + 1 โ 1)/โ(1 โใ ๐กใ^2 ) =(ใ1 โ ๐กใ^2 โ 1)/โ(1 โใ ๐กใ^2 ) =ใ1 โ ๐กใ^2/โ(1 โใ ๐กใ^2 ) โ" " 1/โ(1 โใ ๐กใ^2 ) =โ(1 โใ ๐กใ^2 ) โ" " 1/โ(1 โใ ๐กใ^2 ) Integrating ๐ค.๐.๐ก.๐ฅ โซ1โ(โ ๐ก^2)/(1 โใ ๐กใ^2 ) dt = โซ1โใ(โ(1 โใ ๐กใ^2 ) โ" " 1/โ(1 โใ ๐กใ^2 )) ใ ๐๐ก = โซ1โโ(1^2 โใ ๐กใ^2 ) ๐๐กโโซ1โ1/โ(1^2 โใ ๐กใ^2 ) ๐๐ก = ๐ก/2 โ(1^2 โใ ๐กใ^2 )+1^2/2 sin^(โ1)โกใ๐ก/1ใโsin^(โ1)โกใ๐ก/1ใ We know that โซ1โโ(๐^2โ๐ฅ^2 )=๐ฅ/2 โ(๐^2โ๐ฅ^2 )+๐^2/2 sin^(โ1)โกใ๐ฅ/๐ใ+๐ถ โซ1โ1/โ(๐^2 โ ๐ฅ^2 )=sin^(โ1)โกใ๐ฅ/๐ใ+๐ถ = ๐ก/2 โ(1 โใ ๐กใ^2 )+1/2 sin^(โ1)โก๐กโsin^(โ1)โก๐ก = ๐ก/2 โ(1 โใ ๐กใ^2 ) โ 1/2 sin^(โ1)โก๐ก Hence we can write ๐ผ1 = ๐ก^2 sin^(โ1)โก๐ก+โซ1โ(โ ๐ก^2)/โ(1 โใ ๐กใ^2 ) ๐๐ก ๐ผ1 = ๐ก^2 sinโก๐ก+๐ก/2 โ(1 โใ ๐กใ^2 )โ1/2 sin^(โ1)โก๐ก Putting ๐ก = โ๐ฅ ๐ผ1 = (โ๐ฅ)^2 sinโกโ๐ฅ+โ๐ฅ/2 โ(1 โใ (โ๐ฅ)ใ^2 )โ1/2 sin^(โ1)โกโ๐ฅ ๐ผ1 = ๐ฅ sinโกโ๐ฅ+โ๐ฅ/2 โ(1โ๐ฅ)โ1/2 sin^(โ1)โกโ๐ฅ Hence ๐ผ = 4/๐ ใ ๐ผใ_(1 )โ๐ฅ+C_1 ๐ผ = 4/๐ (๐ฅ ใ๐ ๐๐ใ^(โ1) โ๐ฅ+โ๐ฅ/2 โ(1โ๐ฅ)โ1/2 ใ๐ ๐๐ใ^(โ1) โ๐ฅ)โ๐ฅ+C_1 ๐ผ = 4/๐ (๐ฅ ใ๐ ๐๐ใ^(โ1) โ๐ฅ+โ(๐ฅ โ ๐ฅ^2 )/2 โ1/2 ใ๐ ๐๐ใ^(โ1) โ๐ฅ)โ๐ฅ+C_1 ๐ผ = 4/๐ ๐ฅ ใ๐ ๐๐ใ^(โ1) โ๐ฅ+2/๐ โ(๐ฅ โ ๐ฅ^2 )โ2/๐ ใ๐ ๐๐ใ^(โ1) โ๐ฅโ๐ฅ+C_1 ๐ผ = 4/๐ ๐ฅ ใ๐ ๐๐ใ^(โ1) โ๐ฅโ2/๐ ใ๐ ๐๐ใ^(โ1) โ๐ฅ+2/๐ โ(๐ฅ โ ๐ฅ^2 )โ๐ฅ+C_1 ๐ผ = ใ๐ ๐๐ใ^(โ1) โ๐ฅ [4๐ฅ/๐โ2/๐]+(2 โ(๐ฅ โ ๐ฅ^2 ))/๐โ๐ฅ+ C_1 ๐ผ = ใ๐ ๐๐ใ^(โ1) โ๐ฅ [(4๐ฅ โ 2)/๐]+(2 โ(๐ฅ โ ๐ฅ^2 ))/๐โ๐ฅ+ C_1 ๐ฐ = ใ๐๐๐ใ^(โ๐) โ๐ [(๐(๐๐ โ๐))/๐ ]+(๐ โ(๐ โ ๐^๐ ))/๐ โ๐+ ๐ช_๐