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


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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

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.