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Example 27 (Method 1) Evaluate ∫_0^1▒tan^(−1)⁡𝑥/(1 + 𝑥^2 ) 𝑑𝑥 Step 1 : Let F(𝑥)=∫1▒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, ∫1▒tan^(−1)⁡𝑥/(1+〖 𝑥〗^2 ) 𝑑𝑥=∫1▒〖𝑡/(1+𝑥^2 ) × (1+𝑥^2 )𝑑𝑡〗 =∫1▒〖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 : ∫1▒〖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

  1. Chapter 7 Class 12 Integrals
  2. Serial order wise

About the Author

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo