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Question 1 Important Deleted for CBSE Board 2024 Exams

Question 2 Deleted for CBSE Board 2024 Exams

Question 3 (Supplementary NCERT) Important Deleted for CBSE Board 2024 Exams

Last updated at June 13, 2023 by Teachoo

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