Chapter 7 Class 12 Integrals
Concept wise

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Example 27 (Method 1) Evaluate ∫_0^1β–’tan^(βˆ’1)⁑π‘₯/(1 + π‘₯^2 ) 𝑑π‘₯ Step 1 : Let F(π‘₯)=∫1β–’tan^(βˆ’1)⁑π‘₯/(1+γ€– π‘₯γ€—^2 ) 𝑑π‘₯ Put tan^(βˆ’1)⁑π‘₯=𝑑 Differentiating w.r.t.π‘₯ 𝑑/𝑑π‘₯ (tan^(βˆ’1)⁑π‘₯ )=𝑑𝑑/𝑑π‘₯ 1/(1 + π‘₯^2 )=𝑑𝑑/𝑑π‘₯ Therefore, ∫1β–’tan^(βˆ’1)⁑π‘₯/(1+γ€– π‘₯γ€—^2 ) 𝑑π‘₯=∫1▒〖𝑑/(1+π‘₯^2 ) Γ— (1+π‘₯^2 )𝑑𝑑〗 =∫1▒〖𝑑 𝑑𝑑〗 =𝑑^2/2 Putting 𝑑=γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1)⁑π‘₯ =(tan^(βˆ’1)⁑π‘₯ )^2/2 Hence 𝐹(π‘₯)=(tan^(βˆ’1)⁑π‘₯ )^2/2 Step 2 : ∫1β–’γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1)⁑π‘₯/(1 + π‘₯^2 )=𝐹(1)βˆ’F(0) =1/2 (tan^(βˆ’1)⁑1 )^2 βˆ’1/2 (tan^(βˆ’1)⁑0 )^2 =1/2 (πœ‹/4)^2βˆ’1/2 (0)^2 =1/2 πœ‹^2/16 = 𝝅^𝟐/πŸ‘πŸ Example 27 (Method 2) Evaluate ∫_0^1β–’tan^(βˆ’1)⁑π‘₯/(1 + π‘₯^2 ) 𝑑π‘₯ Put 𝑑=tan^(βˆ’1)⁑π‘₯ Differentiating w.r.t.π‘₯ 𝑑𝑑/𝑑π‘₯=𝑑/𝑑π‘₯ (tan^(βˆ’1)⁑π‘₯ ) 𝑑𝑑/𝑑π‘₯=1/(1 + π‘₯^2 ) (1+π‘₯^2 )𝑑𝑑=𝑑π‘₯ Hence when value of x varies from 0 to 1, value of t varies from 0 to πœ‹/4 Therefore, ∫_0^1β–’tan^(βˆ’1)⁑π‘₯/(1 + π‘₯^2 )=∫_0^(πœ‹/4)▒𝑑/(1 + π‘₯^2 ) 𝑑π‘₯ (1+π‘₯^2 )𝑑𝑑 =∫_0^(πœ‹/4)β–’γ€– 𝑑 𝑑𝑑〗 =[𝑑^2/2]_0^(πœ‹/4) =1/2 [(πœ‹/4)^2βˆ’(0)^2 ] =1/2 Γ— πœ‹^2/16 = 𝝅^𝟐/πŸ‘πŸ

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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, Social Science, Physics, Chemistry, Computer Science at Teachoo.