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

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Transcript

Example 29 (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 29 (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﷯

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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.