Ex 7.2, 20 - Chapter 7 Class 12 Integrals
Last updated at July 11, 2018 by Teachoo
Transcript
Ex 7.2, 20 (π^2π₯ β π^(β2π₯))/(π^2π₯ + π^(β2π₯) ) Step 1: Let π^2π₯ + π^(β2π₯)= π‘ Differentiating both sides π€.π.π‘.π₯ π^2π₯. π(2π₯)/ππ₯ +π^(β2π₯) π(β2π₯)/ππ₯= ππ‘/ππ₯ γ2πγ^2π₯βγ2πγ^(β2π₯)= ππ‘/ππ₯ 2(π^2π₯βπ^(β2π₯) )=ππ‘/ππ₯ " " ππ₯ = ππ‘/2(π^2π₯β π^(β2π₯) ) Step 2: Integrating the function β«1βγ" " (π^2π₯ β π^(β2π₯))/(π^2π₯ + π^(β2π₯) )γ. ππ₯ Putting π^2π₯ + π^(β2π₯)=π‘ & ππ₯=ππ‘/2(π^2π₯β π^(β2π₯) ) = β«1βγ" " (π^2π₯ β π^(β2π₯))/π‘γ. ππ‘/2(π^2π₯β π^(β2π₯) ) = β«1βγ" " 1/2π‘γ. ππ‘ = 1/2 β«1β1/π‘. ππ‘ = 1/2 logβ‘γ |π‘|γ+πΆ = 1/2 logβ‘γ |π^2π₯ + π^(β2π₯) |γ+πΆ = π/π πππβ‘γ (π^ππ + π^(βππ) )γ+πͺ
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