Example 42 - Chapter 7 Class 12 Integrals (Important Question)
Last updated at April 16, 2024 by Teachoo
Chapter 7 Class 12 Integrals
Ex 7.1, 18 Important
Ex 7.1, 20
Ex 7.2, 20 Important
Ex 7.2, 26 Important
Ex 7.2, 35
Ex 7.2, 36 Important
Ex 7.3, 6 Important
Ex 7.3, 13 Important
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Ex 7.3, 22 Important
Ex 7.3, 24 (MCQ) Important
Example 9 (i)
Example 10 (i)
Ex 7.4, 8 Important
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Ex 7.4, 22
Ex 7.4, 25 (MCQ) Important
Example 15 Important
Ex 7.5, 9 Important
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Ex 7.5, 18 Important
Ex 7.5, 21 Important
Example 20 Important
Example 22 Important
Ex 7.6, 13 Important
Ex 7.6, 14 Important
Ex 7.6, 18 Important
Ex 7.6, 19
Ex 7.6, 24 (MCQ) Important
Ex 7.7, 5 Important
Ex 7.7, 10
Ex 7.7, 11 Important
Question 1 Important Deleted for CBSE Board 2024 Exams
Question 4 Important Deleted for CBSE Board 2024 Exams
Question 6 Important Deleted for CBSE Board 2024 Exams
Example 25 (i)
Ex 7.8, 15
Ex 7.8, 16 Important
Ex 7.8, 20 Important
Ex 7.8, 22 (MCQ)
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Ex 7.9, 7 Important
Ex 7.9, 8
Ex 7.9, 9 (MCQ) Important
Example 28 Important
Example 32 Important
Example 34 Important
Ex 7.10,8 Important
Ex 7.10, 18 Important
Example 38 Important
Example 39 Important
Example 42 Important You are here
Misc 18 Important
Misc 8 Important
Question 1 Important Deleted for CBSE Board 2024 Exams
Misc 23 Important
Misc 29 Important
Question 2 Important Deleted for CBSE Board 2024 Exams
Misc 38 (MCQ) Important
Question 4 (MCQ) Important Deleted for CBSE Board 2024 Exams
Integration Formula Sheet - Chapter 41 Class 41 Formulas Important
Chapter 7 Class 12 Integrals
Last updated at April 16, 2024 by Teachoo
Example 42 Evaluate โซ_0^๐โ(๐ฅ ๐๐ฅ)/(๐^2 cos^2โกใ๐ฅ + ๐^2 ใ sin^2โก๐ฅ )Let I= โซ_0^๐โใ๐ฅ/(๐^2 ๐๐๐ ^2 ๐ฅ + ๐^2 ๐ ๐๐^2 ๐ฅ) ๐๐ฅใ โด I=โซ_0^๐โใ((๐ โ ๐ฅ))/(๐^2 ๐๐๐ ^2 (๐ โ ๐ฅ) + ๐^2 ๐ ๐๐^2 (๐ โ ๐ฅ) ) ๐๐ฅใ I=โซ_0^๐โใ(๐ โ ๐ฅ)/(๐^2 [๐๐๐ (๐ โ ๐ฅ)]^2 + ๐^2 [๐ ๐๐(๐ โ ๐ฅ)]^2 ) ๐๐ฅใ I=โซ_0^๐โใ(๐ โ ๐ฅ)/(๐^2 [โ ๐๐๐ ๐ฅ]^2 + ๐^2 [๐ ๐๐ ๐ฅ]^2 ) ๐๐ฅใ I=โซ_0^๐โใ(๐ โ ๐ฅ)/(๐^2 cos^2โก๐ฅ + ๐^2 sin^2โก๐ฅ ) ๐๐ฅใ Adding (1) and (2) i.e. (1) + (2) I+I=โซ_0^๐โใ๐ฅ/(๐^2 cos^2โก๐ฅ + ๐^2 sin^2โก๐ฅ ) ๐๐ฅใ+โซ1โ(๐ โ ๐ฅ)/(๐^2 cos^2โก๐ฅ + ๐^2 sin^2โก๐ฅ ) ๐๐ฅ 2I=โซ_0^๐โ(๐ฅ + ๐ โ ๐ฅ)/(๐^2 cos^2โก๐ฅ + ๐^2 sin^2โก๐ฅ ) ๐๐ฅ 2I=โซ_0^๐โ(๐ )/(๐^2 cos^2โก๐ฅ + ๐^2 sin^2โก๐ฅ ) ๐๐ฅ I=๐/2 โซ_0^๐โใ1/(๐^2 cos^2โก๐ฅ + ๐^2 sin^2โก๐ฅ ) ๐๐ฅใ Dividing numerator and denominator by ๐๐๐ ^2 ๐ฅ, we get I=๐/2 โซ_0^๐โใ(1/cos^2โก๐ฅ )/((๐^2 cos^2โกใ๐ฅ + ๐^2 sin^2โก๐ฅ ใ)/cos^2โก๐ฅ ) ๐๐ฅใ I=๐/2 โซ_0^๐โใ(๐ ๐๐^2 ๐ฅ)/((๐^2 cos^2โก๐ฅ)/cos^2โก๐ฅ + (๐^2 sin^2โก๐ฅ)/cos^2โก๐ฅ ) ๐๐ฅใ I=๐/2 โซ_0^๐โใ(๐ ๐๐^2 ๐ฅ)/(๐^2 + ๐^2 tan^2โก๐ฅ ) ๐๐ฅใ Let ๐(๐ฅ)=sec^2โก๐ฅ/(๐^2 + ๐^2 tan^2โก๐ฅ ) and a = ฯ Now, ๐(2๐โ๐ฅ)=sec^2โก(๐ โ ๐ฅ)/(๐^2 + ๐^2 tan^2โก(๐ โ ๐ฅ) ) ๐(2๐โ๐ฅ)=[โ๐ ๐๐ ๐ฅ]^2/(๐^2 + ๐^2 [โtanโก๐ฅ ]^2 ) ๐(2๐โ๐ฅ)=(๐ ๐๐^2 ๐ฅ)/(๐^2 + ๐^2 tan^2โก๐ฅ ) Therefore, ๐(๐ฅ)=๐(2๐โ๐ฅ) Therefore, I=๐/2 โซ_0^๐โใ(๐ ๐๐^2 ๐ฅ)/(๐^2 + ๐^2 tan^2โก๐ฅ ) ๐๐ฅใ =๐/2 ร 2 โซ_0^(๐/2)โใ(๐ ๐๐^2 ๐ฅ)/(๐^2 + ๐^2 tan^2โก๐ฅ ) ๐๐ฅใ =๐โซ_0^(๐/2)โใ(๐ ๐๐^2 ๐ฅ)/(๐^2 + ๐^2 tan^2โก๐ฅ ) ๐๐ฅใ Let ๐ tanโกใ๐ฅ=๐กใ Differentiating both sides w.r.t. ๐ฅ ๐ ๐ ๐๐^2 ๐ฅ ๐๐ฅ=๐๐ก ๐๐ก=๐๐ก/(๐^2 ๐ ๐๐^2 ๐ฅ) Putting the values of tan ๐ฅ and ๐๐ฅ , we get ๐ผ=๐โซ1_0^(๐/2)โใ(๐ ๐๐^2 ๐ฅ)/(๐^2 + ๐ก^2 ) . ๐๐ฅใ ๐ผ=๐ โซ1_0^โโใ(๐ ๐๐^2 ๐ฅ)/(๐^2 + ๐ก^2 ) .๐๐ก/(๐ ๐ ๐๐^2 ๐ฅ)ใ ๐ผ=๐/๐ โซ1_0^โโ๐๐ก/(๐^2 + ๐ก^2 ) ๐ผ= ใ๐/๐ [1/๐ tan^(โ1)โก(๐ก/๐) ]ใ_0^โ Putting limits, I=๐/๐ [1/๐ ใ๐ก๐๐ใ^(โ1) (โ/๐)โ1/๐ ใ๐ก๐๐ใ^(โ1) (0/๐)] I =๐/๐ [ใ1/๐ tan^(โ1)ใโกใ(โ)โ1/๐ tan^(โ1)โก(0) ใ ] I =๐/๐ (1/๐ (๐/2)โ0) I =๐ ^๐/๐๐๐