Integration Full Chapter Explained - https://you.tube/Integration-Class-12

Last updated at Dec. 23, 2019 by Teachoo

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Misc 21 Integrate the function (2 + sinβ‘2π₯)/(1 + cosβ‘2π₯ ) π^π₯ We can write integral as ((2 +γ sinγβ‘2π₯)/(1 + cosβ‘2π₯ )) π^π₯ = [(2 + sinβ‘2π₯)/(1 + (2 cos^2β‘γπ₯ β 1γ ) )] π^π₯ = [(2 + sinβ‘2π₯)/(2 cos^2β‘π₯ )] π^π₯ = [(2 + 2 cosβ‘γπ₯ sinβ‘π₯ γ)/(2 cos^2β‘π₯ )]ππ₯ = [2(1 +γ cosγβ‘γπ₯ sinβ‘π₯ γ )/(2 cos^2β‘π₯ )]ππ₯ (cosβ‘2π₯=2 cos^2β‘γπ₯β1γ ) (sinβ‘2π₯=2 cosβ‘γπ₯ sinβ‘π₯ γ) = [(1 + cosβ‘γπ₯ sinβ‘π₯ γ)/cos^2β‘π₯ ]ππ₯ = [1/cos^2β‘π₯ +cosβ‘γπ₯ sinβ‘π₯ γ/cos^2β‘π₯ ] π^π₯ = [sec^2β‘γπ₯+cosβ‘γπ₯ sinβ‘π₯ γ/cosβ‘γπ₯ cosβ‘π₯ γ γ ] π^π₯ = [sec^2β‘γπ₯+tanβ‘π₯ γ ] π^π₯ = [tanβ‘γπ₯+sec^2β‘π₯ γ ] π^π₯ It is of the form β«1βγπ^π₯ [π(π₯)+π^β² (π₯)] γ ππ₯=π^π₯ π(π₯)+πΆ Where π(π₯)=tanβ‘π₯ π^β² (π₯)=sec^2β‘π₯ So, our equation becomes β«1βγ[(2 + sinβ‘2π₯)/(1 + cosβ‘2π₯ )] π^π₯ ππ₯=β«1βγπ^π₯ [tanβ‘γπ₯+sec^2β‘π₯ γ ]ππ₯γγ =π^π πππβ‘π+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.