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

Transcript

Ex 7.2, 34 Integrate √(tan⁑π‘₯ )/sin⁑〖π‘₯ cos⁑π‘₯ γ€— Simplifying the function √(tan⁑π‘₯ )/sin⁑〖π‘₯ cos⁑π‘₯ γ€— = √(tan⁑π‘₯ )/(sin⁑〖π‘₯ cos⁑π‘₯ γ€—. cos⁑π‘₯/cos⁑π‘₯ ) = √(tan⁑π‘₯ )/(sin⁑π‘₯ . cos^2⁑π‘₯/cos⁑π‘₯ ) = √(tan⁑π‘₯ )/(cos^2⁑π‘₯ . (sin π‘₯)/cos⁑π‘₯ ) Concept: There are two methods to deal with π‘‘π‘Žπ‘›β‘π‘₯ (1) Convert into 𝑠𝑖𝑛⁑π‘₯ and π‘π‘œπ‘ β‘π‘₯ , then solve using the properties of 𝑠𝑖𝑛⁑π‘₯ and π‘π‘œπ‘ β‘π‘₯ . (2) Change into sec2x, as derivative of tan x is sec2 . Here, 1st Method is not applicable , so we have used 2nd Method . = √(tan⁑π‘₯ )/(cos^2⁑π‘₯ . tan⁑π‘₯ ) = (tan⁑π‘₯ )^(1/2 βˆ’ 1) Γ— 1/cos^2⁑π‘₯ = (tan⁑π‘₯ )^((βˆ’1)/2) Γ— 1/cos^2⁑π‘₯ = (tan⁑π‘₯ )^((βˆ’1)/2) Γ— sec^2⁑π‘₯ ∴ √(tan⁑π‘₯ )/sin⁑〖π‘₯ cos⁑π‘₯ γ€— " = " (tan⁑π‘₯ )^((βˆ’1)/2) " Γ— " sec^2⁑π‘₯ Step 2: Integrating the function ∫1β–’γ€– √(tan⁑π‘₯ )/sin⁑〖π‘₯ cos⁑π‘₯ γ€— γ€— . 𝑑π‘₯ = ∫1β–’γ€– (tan⁑π‘₯ )^((βˆ’1)/2) " Γ— " sec^2⁑π‘₯ γ€—. 𝑑π‘₯" " Let tan⁑π‘₯ = 𝑑 Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯ sec^2⁑π‘₯=𝑑𝑑/𝑑π‘₯ 𝑑π‘₯=𝑑𝑑/sec^2⁑π‘₯ Thus, our equation becomes ∴ ∫1β–’γ€– (tan⁑π‘₯ )^((βˆ’1)/2) " ." sec^2⁑π‘₯ γ€—. 𝑑π‘₯" " = ∫1β–’γ€– (𝑑)^((βˆ’1)/2) " " . sec^2⁑π‘₯ γ€—. 𝑑𝑑/sec^2⁑π‘₯ " " = ∫1▒〖𝑑^((βˆ’1)/2) . 𝑑𝑑〗 = 𝑑^(βˆ’ 1/2 +1)/(βˆ’ 1/2 +1) + 𝐢 = 𝑑^(1/2)/(1/2) + 𝐢 = γ€–2𝑑〗^(1/2)+ 𝐢 = 2βˆšπ‘‘+ 𝐢 = 𝟐√(π­πšπ§β‘π’™ )+ π‘ͺ (Using 𝑑=π‘‘π‘Žπ‘›β‘π‘₯)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.