Calculus 2: Integration of Trig Functions (5 of 16) Integral of cotx=?
Calculus 2: Integration of Trig Functions (5 of 16) Integral of cotx=?

Slide32.JPG
Slide33.JPG
Slide34.JPG
Slide35.JPG
Slide36.JPG
Slide37.JPG

Examples

Example 1 (ii)

Example 1 (iii)

Example 2 (i)

Example 2 (ii)

Example 2 (iii) Important

Example 3 (i)

Example 3 (ii) Important

Example 3 (iii)

Example 4

Example 5 (i)

Example 5 (ii)

Example 5 (iii) Important

Example 5 (iv) Important

Example 6 (i)

Example 6 (ii) Important

Example 6 (iii) Important

Example 7 (i)

Example 7 (ii) Important

Example 7 (iii)

Example 8 (i)

Example 8 (ii) Important

Example 9 (i)

Example 9 (ii) Important

Example 9 (iii) Important

Example 10 (i)

Example 10 (ii) Important

Example 11

Example 12

Example 13 Important

Example 14

Example 15 Important

Example 16 Important

Example 17

Example 18 Important

Example 19

Example 20 Important

Example 21 Important

Example 22 Important

Example 23

Example 24

Example 25 (i)

Example 25 (ii) Important

Example 25 (iii)

Example 25 (iv) Important

Example 26

Example 27

Example 28 Important

Example 29

Example 30

Example 31

Example 32 Important

Example 33 Important

Example 34 Important

Example 35

Example 36 Important

Example 37 Important

Example 38 Important

Example 39 Important You are here

Example 40 Important

Example 41 Important

Example 42 Important

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 39 Evaluate ∫1▒[√(cot𝑥 )+√(tan𝑥 )] 𝑑𝑥 ∫1▒[√(cot𝑥 )+√(tan𝑥 )] 𝑑𝑥 =∫1▒[√(cot𝑥 )+1/√(cot𝑥 )] 𝑑𝑥 =∫1▒[(cot𝑥 + 1)/√(cot𝑥 )] 𝑑𝑥 =∫1▒[√(tan𝑥 ) (cot𝑥+1)] 𝑑𝑥 Let tan𝑥=𝑡^2 Differentiating both sides 𝑤.𝑟.𝑡.𝑥. sec^2 𝑥=2𝑡 𝑑𝑡/𝑑𝑥 1+tan^2 𝑥=2𝑡 . 𝑑𝑡/𝑑𝑥 1+(𝑡^2 )^2=2𝑡 . 𝑑𝑡/𝑑𝑥 1+𝑡^4=2𝑡 . 𝑑𝑡/𝑑𝑥 (1+𝑡^4 ) 𝑑𝑥=2𝑡 𝑑𝑡 𝑑𝑥=2𝑡/(1 + 𝑡^4 ) . 𝑑𝑡 Putting values of t & dt, we get ∫1▒[√(tan𝑥 ) (cot𝑥+1)] 𝑑𝑥 = ∫1▒[√(𝑡^2 ) (cot𝑥+1)] 𝑑𝑥 = ∫1▒[√(𝑡^2 ) (1/tan𝑥 +1)] 𝑑𝑥 = ∫1▒𝑡[1/𝑡^2 +1] 𝑑𝑥 = ∫1▒𝑡[(1 + 𝑡^2)/𝑡^2 ] ×2𝑡/(1 + 𝑡^4 ) . 𝑑𝑡 = ∫1▒2[(1 + 𝑡^2)/(1 + 𝑡^4 )] 𝑑𝑡 = 2∫1▒(1 + 𝑡^2)/(1 + 𝑡^4 ) 𝑑𝑡 Dividing numerator and denominator by 𝑡^2 = 2 ∫1▒((1 + 𝑡^2)/𝑡^2 )/((1 + 𝑡^4)/𝑡^2 ) . 𝑑𝑡 = 2 ∫1▒(1/𝑡^2 + 1)/(1/𝑡^2 + 𝑡^2 ) . 𝑑𝑡 = 2 ∫1▒(1/𝑡^2 + 1)/(1/𝑡^2 + 𝑡^2 ) . 𝑑𝑡 = 2 ∫1▒(1 + 1/𝑡^2 )/( 𝑡^2 + 1/𝑡^2 + 2 − 2) . 𝑑𝑡 = 2 ∫1▒(1 + 1/𝑡^2 )/( (𝑡)^2 + (1/𝑡)^2− 2 (𝑡) (1/𝑡) + 2) . 𝑑𝑡 = 2 ∫1▒(1 + 1/𝑡^2 )/((𝑡 − 1/𝑡)^2 + 2) . 𝑑𝑡 = 2 ∫1▒(1 + 1/𝑡^2 )/((𝑡 − 1/𝑡)^2 +(√2 )^2 ) . 𝑑𝑡 Let 𝑡−1/𝑡=𝑦 Differentiating both sides 𝑤.𝑟.𝑡.𝑥. 1+ 1/𝑡^2 = 𝑑𝑦/𝑑𝑡 𝑑𝑡 =𝑑𝑦/((1 + 1/𝑡^2 ) ) Putting the values of (1/t −t) and dt, we get = 2 ∫1▒(1 + 1/𝑡^2 )/(𝑦^2 +(√2 )^2 ) . 𝑑𝑡 = 2 ∫1▒((1 + 1/𝑡^2 ))/(𝑦^2 + (√2 )^2 ) × 𝑑𝑦/((1 + 1/𝑡^2 ) ) = 2 ∫1▒1/(𝑦^2 + (√2 )^2 ) . 𝑑𝑦 = 2(1/√2 tan^(−1)〖 𝑦/√2〗 +𝐶1) = 2/√2 tan^(−1)〖 𝑦/√2〗 +2𝐶1 = √2 tan^(−1)〖 (1/𝑡 − 𝑡)/√2〗 +2𝐶1 = √2 tan^(−1)〖 (𝑡^2 − 1)/(√2 𝑡)〗 +𝐶 = √2 tan^(−1)((tan𝑥 − 1)/(√2 √(tan𝑥 )))+𝐶 = √𝟐 〖𝒕𝒂𝒏〗^(−𝟏)((𝒕𝒂𝒏𝒙 − 𝟏)/(√(𝟐 𝒕𝒂𝒏𝒙 ) ))+𝑪

You are watching: Evaluate integral [root cot x + root tan x] dx. Info created by Bút Chì Xanh selection and synthesis along with other related topics.