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Ex 7.2, 34 Integrate √(tan⁡𝑥 )/sin⁡〖𝑥 cos⁡𝑥 〗 Simplifying the function √(tan⁡𝑥 )/sin⁡〖𝑥 cos⁡𝑥 〗 = √(tan⁡𝑥 )/(sin⁡〖𝑥 cos⁡𝑥 〗. cos⁡𝑥/cos⁡𝑥 ) = √(tan⁡𝑥 )/(sin⁡𝑥 . cos^2⁡𝑥/cos⁡𝑥 ) = √(tan⁡𝑥 )/(cos^2⁡𝑥 . (sin 𝑥)/cos⁡𝑥 ) Concept: There are two methods to deal with 𝑡𝑎𝑛⁡𝑥 (1) Convert into 𝑠𝑖𝑛⁡𝑥 and 𝑐𝑜𝑠⁡𝑥 , then solve using the properties of 𝑠𝑖𝑛⁡𝑥 and 𝑐𝑜𝑠⁡𝑥 . (2) Change into sec2x, as derivative of tan x is sec2 . Here, 1st Method is not applicable , so we have used 2nd Method . = √(tan⁡𝑥 )/(cos^2⁡𝑥 . tan⁡𝑥 ) = (tan⁡𝑥 )^(1/2 − 1) × 1/cos^2⁡𝑥 = (tan⁡𝑥 )^((−1)/2) × 1/cos^2⁡𝑥 = (tan⁡𝑥 )^((−1)/2) × sec^2⁡𝑥 ∴ √(tan⁡𝑥 )/sin⁡〖𝑥 cos⁡𝑥 〗 " = " (tan⁡𝑥 )^((−1)/2) " × " sec^2⁡𝑥 Step 2: Integrating the function ∫1▒〖 √(tan⁡𝑥 )/sin⁡〖𝑥 cos⁡𝑥 〗 〗 . 𝑑𝑥 = ∫1▒〖 (tan⁡𝑥 )^((−1)/2) " × " sec^2⁡𝑥 〗. 𝑑𝑥" " Let tan⁡𝑥 = 𝑡 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 sec^2⁡𝑥=𝑑𝑡/𝑑𝑥 𝑑𝑥=𝑑𝑡/sec^2⁡𝑥 Thus, our equation becomes ∴ ∫1▒〖 (tan⁡𝑥 )^((−1)/2) " ." sec^2⁡𝑥 〗. 𝑑𝑥" " = ∫1▒〖 (𝑡)^((−1)/2) " " . sec^2⁡𝑥 〗. 𝑑𝑡/sec^2⁡𝑥 " " = ∫1▒〖𝑡^((−1)/2) . 𝑑𝑡〗 = 𝑡^(− 1/2 +1)/(− 1/2 +1) + 𝐶 = 𝑡^(1/2)/(1/2) + 𝐶 = 〖2𝑡〗^(1/2)+ 𝐶 = 2√𝑡+ 𝐶 = 𝟐√(𝐭𝐚𝐧⁡𝒙 )+ 𝑪 (Using 𝑡=𝑡𝑎𝑛⁡𝑥)

  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