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Example 39 Evaluate ∫1▒[√(cot⁡𝑥 )+√(tan⁡𝑥 )] 𝑑𝑥 ∫1▒[√(cot⁡𝑥 )+√(tan⁡𝑥 )] 𝑑𝑥 =∫1▒[√(cot⁡𝑥 )+1/√(cot⁡𝑥 )] 𝑑𝑥 =∫1▒[(cot⁡𝑥 + 1)/√(cot⁡𝑥 )] 𝑑𝑥 =∫1▒[√(tan⁡𝑥 ) (cot⁡𝑥+1)] 𝑑𝑥 Let tan⁡𝑥=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▒[√(tan⁡𝑥 ) (cot⁡𝑥+1)] 𝑑𝑥 = ∫1▒[√(𝑡^2 ) (cot⁡𝑥+1)] 𝑑𝑥 = ∫1▒[√(𝑡^2 ) (1/tan⁡𝑥 +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)⁡((tan⁡𝑥 − 1)/(√2 √(tan⁡𝑥 )))+𝐶 = √𝟐 〖𝒕𝒂𝒏〗^(−𝟏)⁡((𝒕𝒂𝒏⁡𝒙 − 𝟏)/(√(𝟐 𝒕𝒂𝒏⁡𝒙 ) ))+𝑪

  1. Chapter 7 Class 12 Integrals
  2. Serial order wise

About the Author

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo