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Misc 21 - Integarate 2 + sin 2x / 1 + cos 2x ex - Miscellaneous

Misc 21 - Chapter 7 Class 12 Integrals - Part 2
Misc 21 - Chapter 7 Class 12 Integrals - Part 3

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Misc 21 Integrate the function (2 + sin⁑2π‘₯)/(1 + cos⁑2π‘₯ ) 𝑒^π‘₯ We can write integral as ((2 +γ€– sin〗⁑2π‘₯)/(1 + cos⁑2π‘₯ )) 𝑒^π‘₯ = [(2 + sin⁑2π‘₯)/(1 + (2 cos^2⁑〖π‘₯ βˆ’ 1γ€— ) )] 𝑒^π‘₯ = [(2 + sin⁑2π‘₯)/(2 cos^2⁑π‘₯ )] 𝑒^π‘₯ = [(2 + 2 cos⁑〖π‘₯ sin⁑π‘₯ γ€—)/(2 cos^2⁑π‘₯ )]𝑒π‘₯ = [2(1 +γ€– cos〗⁑〖π‘₯ sin⁑π‘₯ γ€— )/(2 cos^2⁑π‘₯ )]𝑒π‘₯ (cos⁑2π‘₯=2 cos^2⁑〖π‘₯βˆ’1γ€— ) (sin⁑2π‘₯=2 cos⁑〖π‘₯ sin⁑π‘₯ γ€—) = [(1 + cos⁑〖π‘₯ sin⁑π‘₯ γ€—)/cos^2⁑π‘₯ ]𝑒π‘₯ = [1/cos^2⁑π‘₯ +cos⁑〖π‘₯ sin⁑π‘₯ γ€—/cos^2⁑π‘₯ ] 𝑒^π‘₯ = [sec^2⁑〖π‘₯+cos⁑〖π‘₯ sin⁑π‘₯ γ€—/cos⁑〖π‘₯ cos⁑π‘₯ γ€— γ€— ] 𝑒^π‘₯ = [sec^2⁑〖π‘₯+tan⁑π‘₯ γ€— ] 𝑒^π‘₯ = [tan⁑〖π‘₯+sec^2⁑π‘₯ γ€— ] 𝑒^π‘₯ It is of the form ∫1▒〖𝑒^π‘₯ [𝑓(π‘₯)+𝑓^β€² (π‘₯)] γ€— 𝑑π‘₯=𝑒^π‘₯ 𝑓(π‘₯)+𝐢 Where 𝑓(π‘₯)=tan⁑π‘₯ 𝑓^β€² (π‘₯)=sec^2⁑π‘₯ So, our equation becomes ∫1β–’γ€–[(2 + sin⁑2π‘₯)/(1 + cos⁑2π‘₯ )] 𝑒^π‘₯ 𝑑π‘₯=∫1▒〖𝑒^π‘₯ [tan⁑〖π‘₯+sec^2⁑π‘₯ γ€— ]𝑑π‘₯γ€—γ€— =𝒆^𝒙 𝒕𝒂𝒏⁑𝒙+𝐂

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.