Learn Intergation from Davneet Sir - Live lectures starting soon!
Example 29 - Chapter 7 Class 12 Integrals (Term 2)
Last updated at May 29, 2018 by Teachoo
Examples
Example 1 (i)
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
Example 23
Example 24
Example 25 Important Deleted for CBSE Board 2023 Exams
Example 26 Deleted for CBSE Board 2023 Exams
Example 27 (i)
Example 27 (ii) Important
Example 27 (iii)
Example 27 (iv) Important
Example 28
Example 29 You are here
Example 30 Important
Example 31
Example 32
Example 33
Example 34 Important
Example 35 Important
Example 36 Important
Example 37
Example 38 Important
Example 39 Important
Example 40 Important
Example 41 Important
Example 42 Important
Example 43 Important
Example 44 Important
Example 25 (Supplementary NCERT) Important Deleted for CBSE Board 2023 Exams
Chapter 7 Class 12 Integrals
Serial order wise
Transcript
Example 29 (Method 1) Evaluate 01tan−1𝑥1 + 𝑥2 𝑑𝑥 Step 1 : Let F𝑥=tan−1𝑥1+ 𝑥2𝑑𝑥 Put tan−1𝑥=𝑡 Differentiating w.r.t.𝑥 𝑑𝑑𝑥tan−1𝑥=𝑑𝑡𝑑𝑥 11 + 𝑥2=𝑑𝑡𝑑𝑥 Therefore, tan−1𝑥1+ 𝑥2𝑑𝑥=𝑡1+𝑥2 × 1+𝑥2𝑑𝑡 =𝑡 𝑑𝑡 =𝑡22 Putting 𝑡=𝑡𝑎𝑛−1𝑥 =tan−1𝑥22 Hence 𝐹𝑥=tan−1𝑥22 Step 2 : 𝑡𝑎𝑛−1𝑥1 + 𝑥2=𝐹1−F0 =12tan−112 −12tan−102 =12𝜋42−1202 =12 𝜋216 = 𝝅𝟐𝟑𝟐 Example 29 (Method 2) Evaluate 01tan−1𝑥1 + 𝑥2 𝑑𝑥 Put 𝑡=tan−1𝑥 Differentiating w.r.t.𝑥 𝑑𝑡𝑑𝑥=𝑑𝑑𝑥tan−1𝑥 𝑑𝑡𝑑𝑥=11 + 𝑥2 1+𝑥2𝑑𝑡=𝑑𝑥 Hence when value of x varies from 0 to 1, value of t varies from 0 to 𝜋4 Therefore, 01tan−1𝑥1 + 𝑥2=0𝜋4𝑡1 + 𝑥2𝑑𝑥 1+𝑥2𝑑𝑡 =0𝜋4 𝑡 𝑑𝑡 =𝑡220𝜋4 =12𝜋42−02 =12 × 𝜋216 = 𝝅𝟐𝟑𝟐
