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Ex 7.2, 34 - Integrate root(tan x) / sin x cos x - teachoo

Ex 7.2, 34 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.2, 34 - Chapter 7 Class 12 Integrals - Part 3 Ex 7.2, 34 - Chapter 7 Class 12 Integrals - Part 4

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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 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.