Check sibling questions

Example 41 - Evaluate integral [root cot x + root tan x] dx

Example 41 - Chapter 7 Class 12 Integrals - Part 2
Example 41 - Chapter 7 Class 12 Integrals - Part 3 Example 41 - Chapter 7 Class 12 Integrals - Part 4 Example 41 - Chapter 7 Class 12 Integrals - Part 5 Example 41 - Chapter 7 Class 12 Integrals - Part 6 Example 41 - Chapter 7 Class 12 Integrals - Part 7

Get live Maths 1-on-1 Classs - Class 6 to 12


Transcript

Example 41 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 ∫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 ("Using" π‘‘π‘Žπ‘›β‘π‘₯= 1/π‘π‘œπ‘‘β‘π‘₯ ) 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‘₯ ["Using" π‘‘π‘Žπ‘›β‘π‘₯= 𝑑^2 ] = ∫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 ["Using" π‘‘π‘Žπ‘›β‘π‘₯= 𝑑^2 ] = 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 (Adding and subtracting 2 in denominator) = 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) It is of form ∫1β–’1/(π‘₯^2 +γ€– π‘Žγ€—^2 ) . 𝑑π‘₯ = 1/π‘Ž tan^(βˆ’1)⁑〖 π‘₯/π‘Žγ€— +𝐢1 Replacing x by y and a by √2 we, get = 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 ["Using" 𝑦=1/𝑑 βˆ’π‘‘] [β–ˆ("Using" 𝑑= √(tan⁑π‘₯ ))]

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

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.