Chapter 7 Class 12 Integrals

Ex 7.1, 10 Important

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

Ex 7.3, 18 Important

Ex 7.3, 22 Important

Ex 7.3, 24 (MCQ) Important

Example 9 (i)

Example 10 (i)

Ex 7.4, 8 Important

Ex 7.4, 15 Important

Ex 7.4, 21 Important

Ex 7.4, 22

Ex 7.4, 25 (MCQ) Important

Example 15 Important

Ex 7.5, 9 Important

Ex 7.5, 11 Important

Ex 7.5, 17

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)

Ex 7.9, 4

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 You are here

Example 42 Important

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

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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Example 39 Evaluate ∫1β–’[√(cot⁑π‘₯ )+√(tan⁑π‘₯ )] 𝑑π‘₯ ∫1β–’[√(cot⁑π‘₯ )+√(tan⁑π‘₯ )] 𝑑π‘₯ =∫1β–’[√(cot⁑π‘₯ )+1/√(cot⁑π‘₯ )] 𝑑π‘₯ =∫1β–’[(cot⁑π‘₯ + 1)/√(cot⁑π‘₯ )] 𝑑π‘₯ =∫1β–’[√(tan⁑π‘₯ ) (cot⁑π‘₯+1)] 𝑑π‘₯ Let tan⁑π‘₯=𝑑^2 Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯. sec^2 π‘₯=2𝑑 𝑑𝑑/𝑑π‘₯ 1+tan^2 π‘₯=2𝑑 . 𝑑𝑑/𝑑π‘₯ 1+(𝑑^2 )^2=2𝑑 . 𝑑𝑑/𝑑π‘₯ 1+𝑑^4=2𝑑 . 𝑑𝑑/𝑑π‘₯ (1+𝑑^4 ) 𝑑π‘₯=2𝑑 𝑑𝑑 𝑑π‘₯=2𝑑/(1 + 𝑑^4 ) . 𝑑𝑑 Putting values of t & dt, we get ∫1β–’[√(tan⁑π‘₯ ) (cot⁑π‘₯+1)] 𝑑π‘₯ = ∫1β–’[√(𝑑^2 ) (cot⁑π‘₯+1)] 𝑑π‘₯ = ∫1β–’[√(𝑑^2 ) (1/tan⁑π‘₯ +1)] 𝑑π‘₯ = ∫1▒𝑑[1/𝑑^2 +1] 𝑑π‘₯ = ∫1▒𝑑[(1 + 𝑑^2)/𝑑^2 ] Γ—2𝑑/(1 + 𝑑^4 ) . 𝑑𝑑 = ∫1β–’2[(1 + 𝑑^2)/(1 + 𝑑^4 )] 𝑑𝑑 = 2∫1β–’(1 + 𝑑^2)/(1 + 𝑑^4 ) 𝑑𝑑 Dividing numerator and denominator by 𝑑^2 = 2 ∫1β–’((1 + 𝑑^2)/𝑑^2 )/((1 + 𝑑^4)/𝑑^2 ) . 𝑑𝑑 = 2 ∫1β–’(1/𝑑^2 + 1)/(1/𝑑^2 + 𝑑^2 ) . 𝑑𝑑 = 2 ∫1β–’(1/𝑑^2 + 1)/(1/𝑑^2 + 𝑑^2 ) . 𝑑𝑑 = 2 ∫1β–’(1 + 1/𝑑^2 )/( 𝑑^2 + 1/𝑑^2 + 2 βˆ’ 2) . 𝑑𝑑 = 2 ∫1β–’(1 + 1/𝑑^2 )/( (𝑑)^2 + (1/𝑑)^2βˆ’ 2 (𝑑) (1/𝑑) + 2) . 𝑑𝑑 = 2 ∫1β–’(1 + 1/𝑑^2 )/((𝑑 βˆ’ 1/𝑑)^2 + 2) . 𝑑𝑑 = 2 ∫1β–’(1 + 1/𝑑^2 )/((𝑑 βˆ’ 1/𝑑)^2 +(√2 )^2 ) . 𝑑𝑑 Let π‘‘βˆ’1/𝑑=𝑦 Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯. 1+ 1/𝑑^2 = 𝑑𝑦/𝑑𝑑 𝑑𝑑 =𝑑𝑦/((1 + 1/𝑑^2 ) ) Putting the values of (1/t βˆ’t) and dt, we get = 2 ∫1β–’(1 + 1/𝑑^2 )/(𝑦^2 +(√2 )^2 ) . 𝑑𝑑 = 2 ∫1β–’((1 + 1/𝑑^2 ))/(𝑦^2 + (√2 )^2 ) Γ— 𝑑𝑦/((1 + 1/𝑑^2 ) ) = 2 ∫1β–’1/(𝑦^2 + (√2 )^2 ) . 𝑑𝑦 = 2(1/√2 tan^(βˆ’1)⁑〖 𝑦/√2γ€— +𝐢1) = 2/√2 tan^(βˆ’1)⁑〖 𝑦/√2γ€— +2𝐢1 = √2 tan^(βˆ’1)⁑〖 (1/𝑑 βˆ’ 𝑑)/√2γ€— +2𝐢1 = √2 tan^(βˆ’1)⁑〖 (𝑑^2 βˆ’ 1)/(√2 𝑑)γ€— +𝐢 = √2 tan^(βˆ’1)⁑((tan⁑π‘₯ βˆ’ 1)/(√2 √(tan⁑π‘₯ )))+𝐢 = √𝟐 〖𝒕𝒂𝒏〗^(βˆ’πŸ)⁑((𝒕𝒂𝒏⁑𝒙 βˆ’ 𝟏)/(√(𝟐 𝒕𝒂𝒏⁑𝒙 ) ))+π‘ͺ

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.