Check sibling questions

Example 29 - Evaluate tan-1 x / 1 + x2 dx - Chapter 7 - Definate Integration - By Substitution

Example 29 - Chapter 7 Class 12 Integrals - Part 2
Example 29 - Chapter 7 Class 12 Integrals - Part 3 Example 29 - Chapter 7 Class 12 Integrals - Part 4 Example 29 - Chapter 7 Class 12 Integrals - Part 5

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Example 27 (Method 1) Evaluate ﷐0﷮1﷮﷐﷐﷐tan﷮−1﷯﷮𝑥﷯﷮1 + ﷐𝑥﷮2﷯﷯﷯ 𝑑𝑥 Step 1 : Let F﷐𝑥﷯=﷐﷮﷮﷐﷐﷐tan﷮−1﷯﷮𝑥﷯﷮1+﷐ 𝑥﷮2﷯﷯﷯𝑑𝑥 Put ﷐﷐tan﷮−1﷯﷮𝑥﷯=𝑡 Differentiating w.r.t.𝑥 ﷐𝑑﷮𝑑𝑥﷯﷐﷐﷐tan﷮−1﷯﷮𝑥﷯﷯=﷐𝑑𝑡﷮𝑑𝑥﷯ ﷐1﷮1 + ﷐𝑥﷮2﷯﷯=﷐𝑑𝑡﷮𝑑𝑥﷯ Therefore, ﷐﷮﷮﷐﷐﷐tan﷮−1﷯﷮𝑥﷯﷮1+﷐ 𝑥﷮2﷯﷯﷯𝑑𝑥=﷐﷮﷮﷐𝑡﷮1+﷐𝑥﷮2﷯﷯ × ﷐1+﷐𝑥﷮2﷯﷯𝑑𝑡﷯ =﷐﷮﷮𝑡 𝑑𝑡﷯ =﷐﷐𝑡﷮2﷯﷮2﷯ Putting 𝑡=﷐﷐𝑡𝑎𝑛﷮−1﷯﷮𝑥﷯ =﷐﷐﷐﷐﷐tan﷮−1﷯﷮𝑥﷯﷯﷮2﷯﷮2﷯ Hence 𝐹﷐𝑥﷯=﷐﷐﷐﷐﷐tan﷮−1﷯﷮𝑥﷯﷯﷮2﷯﷮2﷯ Step 2 : ﷐﷮﷮﷐﷐﷐𝑡𝑎𝑛﷮−1﷯﷮𝑥﷯﷮1 + ﷐𝑥﷮2﷯﷯﷯=𝐹﷐1﷯−F﷐0﷯ =﷐1﷮2﷯﷐﷐﷐﷐tan﷮−1﷯﷮1﷯﷯﷮2﷯ −﷐1﷮2﷯﷐﷐﷐﷐tan﷮−1﷯﷮0﷯﷯﷮2﷯ =﷐1﷮2﷯﷐﷐﷐𝜋﷮4﷯﷯﷮2﷯−﷐1﷮2﷯﷐﷐0﷯﷮2﷯ =﷐1﷮2﷯ ﷐﷐𝜋﷮2﷯﷮16﷯ = ﷐﷐𝝅﷮𝟐﷯﷮𝟑𝟐﷯ Example 27 (Method 2) Evaluate ﷐0﷮1﷮﷐﷐﷐tan﷮−1﷯﷮𝑥﷯﷮1 + ﷐𝑥﷮2﷯﷯﷯ 𝑑𝑥 Put 𝑡=﷐﷐tan﷮−1﷯﷮𝑥﷯ Differentiating w.r.t.𝑥 ﷐𝑑𝑡﷮𝑑𝑥﷯=﷐𝑑﷮𝑑𝑥﷯﷐﷐﷐tan﷮−1﷯﷮𝑥﷯﷯ ﷐𝑑𝑡﷮𝑑𝑥﷯=﷐1﷮1 + ﷐𝑥﷮2﷯﷯ ﷐1+﷐𝑥﷮2﷯﷯𝑑𝑡=𝑑𝑥 Hence when value of x varies from 0 to 1, value of t varies from 0 to ﷐𝜋﷮4﷯ Therefore, ﷐0﷮1﷮﷐﷐﷐tan﷮−1﷯﷮𝑥﷯﷮1 + ﷐𝑥﷮2﷯﷯﷯=﷐0﷮﷐𝜋﷮4﷯﷮﷐𝑡﷮1 + ﷐𝑥﷮2﷯﷯﷯𝑑𝑥 ﷐1+﷐𝑥﷮2﷯﷯𝑑𝑡 =﷐0﷮﷐𝜋﷮4﷯﷮ 𝑡 𝑑𝑡﷯ =﷐﷐﷐﷐𝑡﷮2﷯﷮2﷯﷯﷮0﷮﷐𝜋﷮4﷯﷯ =﷐1﷮2﷯﷐﷐﷐﷐𝜋﷮4﷯﷯﷮2﷯−﷐﷐0﷯﷮2﷯﷯ =﷐1﷮2﷯ × ﷐﷐𝜋﷮2﷯﷮16﷯ = ﷐﷐𝝅﷮𝟐﷯﷮𝟑𝟐﷯

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.