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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Misc 20 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‘₯ = [(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

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.