Example 22 - Chapter 7 Class 12 Integrals
Last updated at April 16, 2024 by Teachoo
Examples
Example 1 (ii)
Example 1 (iii)
Example 2 (i)
Example 2 (ii)
Example 2 (iii) Important
Example 3 (i)
Example 3 (ii) Important
Example 3 (iii)
Example 4
Example 5 (i)
Example 5 (ii)
Example 5 (iii) Important
Example 5 (iv) Important
Example 6 (i)
Example 6 (ii) Important
Example 6 (iii) Important
Example 7 (i)
Example 7 (ii) Important
Example 7 (iii)
Example 8 (i)
Example 8 (ii) Important
Example 9 (i)
Example 9 (ii) Important
Example 9 (iii) Important
Example 10 (i)
Example 10 (ii) Important
Example 11
Example 12
Example 13 Important
Example 14
Example 15 Important
Example 16 Important
Example 17
Example 18 Important
Example 19
Example 20 Important
Example 21 Important
Example 22 Important You are here
Example 23
Example 24
Example 25 (i)
Example 25 (ii) Important
Example 25 (iii)
Example 25 (iv) Important
Example 26
Example 27
Example 28 Important
Example 29
Example 30
Example 31
Example 32 Important
Example 33 Important
Example 34 Important
Example 35
Example 36 Important
Example 37 Important
Example 38 Important
Example 39 Important
Example 40 Important
Example 41 Important
Example 42 Important
Question 1 Important
Question 2
Question 3 (Supplementary NCERT) Important
Last updated at April 16, 2024 by Teachoo
Example 22 Find (i) โซ1โ๐^๐ฅ (tan^(โ1)โก๐ฅ+ 1/(1 + ๐ฅ^2 )) ๐๐ฅ โซ1โใ๐^๐ฅ (tan^(โ1)โก๐ฅ+1/(1 + ๐ฅ^2 ))๐๐ฅใ It is of the form โซ1โใ๐^๐ฅ [๐(๐ฅ)+๐^โฒ (๐ฅ)] ใ ๐๐ฅ=๐^๐ฅ ๐(๐ฅ)+๐ถ Where ๐(๐ฅ)=tan^(โ1)โก๐ฅ ๐^โฒ (๐ฅ)= 1/(1 + ๐ฅ^2 ) So, our equation becomes โซ1โใ๐^๐ฅ (tan^(โ1)โก๐ฅ+1/(1 + ๐ฅ^2 ))๐๐ฅใ=๐^๐ ใ๐ญ๐๐งใ^(โ๐)โกใ๐+๐ชใ Example 22 Find (ii) โซ1โ((๐ฅ^2 + 1) ๐^๐ฅ)/(๐ฅ + 1)^2 ๐๐ฅ โซ1โใ(๐ฅ^2 + 1)/(๐ฅ + 1)^2 .๐^๐ฅ ๐๐ฅใ Adding and subtracting 1 in numerator =โซ1โใ(๐ฅ^2+ 1 + 1 โ 1)/(๐ฅ + 1)^2 .๐^๐ฅ .๐๐ฅใ =โซ1โใ(๐ฅ^2 โ 1 + 1 + 1)/(๐ฅ + 1)^2 .๐^๐ฅ .๐๐ฅใ =โซ1โใ[(๐ฅ^2 โ 1)/(๐ฅ + 1)^2 +2/(๐ฅ + 1)^2 ] ๐^๐ฅ ๐๐ฅใ =โซ1โใ๐^๐ฅ [(๐ฅ^2 โ (1)^2)/(๐ฅ + 1)^2 +2/(๐ฅ + 1)^2 ]๐๐ฅใ =โซ1โใ๐^๐ฅ [(๐ฅ โ 1)(๐ฅ + 1)/(๐ฅ + 1)^2 +2/(๐ฅ + 1)^2 ]๐๐ฅใ =โซ1โใ๐^๐ฅ [(๐ฅ โ 1)/(๐ฅ + 1)+2/(๐ฅ + 1)^2 ]๐๐ฅใ It is of form โซ1โใ๐^๐ฅ [๐(๐ฅ)+๐^โฒ (๐ฅ)] ใ ๐๐ฅ=๐^๐ฅ ๐(๐ฅ)+๐ถ Where ๐(๐ฅ)=(๐ฅ โ 1)/(๐ฅ + 1) ๐^โฒ (๐ฅ)=๐/๐๐ฅ [(๐ฅ โ 1)/(๐ฅ + 1)] ๐^โฒ (๐ฅ)=(1.(๐ฅ + 1) โ1 (๐ฅ โ 1))/(๐ฅ + 1)^2 =(๐ฅ + 1 โ ๐ฅ + 1)/(๐ฅ + 1)^2 =2/(๐ฅ + 1)^2 Thus, our equation becomes โซ1โใ(๐ฅ^2 + 1)/(๐ฅ + 1)^2 .๐^๐ฅ=โซ1โใ๐^๐ฅ [(๐ฅ โ 1)/(๐ฅ + 1)+2/(๐ฅ + 1)^2 ]๐๐ฅใใ =๐^๐ฅ [(๐ฅ โ 1)/(๐ฅ + 1)]+๐ถ =(๐ โ ๐)/(๐ + ๐).๐^๐+๐ช