# Ex 7.3, 14

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 7.3, 14 cosβ‘γπ₯ β sinβ‘π₯ γ/(1 + sinβ‘2π₯ ) β«1βcosβ‘γπ₯ β sinβ‘π₯ γ/(1 + sinβ‘2π₯ ) ππ₯ =β«1βcosβ‘γπ₯ βγ sinγβ‘π₯ γ/(1 + 2 sinβ‘π₯ cosβ‘π₯ ) ππ₯ =β«1βcosβ‘γπ₯ βγ sinγβ‘π₯ γ/(sin^2β‘π₯ + cos^2β‘π₯ + 2 sinβ‘cosβ‘π₯ ) ππ₯ =β«1βcosβ‘γπ₯ βγ sinγβ‘π₯ γ/(sinβ‘π₯ + cosβ‘π₯ )^2 ππ₯ Let sinβ‘π₯+cosβ‘π₯=π‘ Differentiating w.r.t.x π(sinβ‘π₯ + cosβ‘π₯ )/ππ₯=ππ‘/ππ₯ cos π₯βsinβ‘π₯=ππ‘/ππ₯ ππ₯=ππ‘/(cos π₯ β sinβ‘π₯ ) Thus, our equation becomes β«1β((cosβ‘γπ₯ β sinβ‘π₯ γ ))/(sinβ‘π₯ + cosβ‘π₯ )^2 ππ₯ =β«1β((cosβ‘γπ₯ β sinβ‘π₯ γ ))/π‘^2 Γππ‘/cosβ‘γπ₯ β sinβ‘π₯ γ =β«1βππ‘/π‘^2 =β«1βπ‘^(β2) ππ‘ =π‘^(β2 +1)/(β2 + 1) +πΆ =π‘^(β1)/(β1) +πΆ =(β1)/π‘ +πΆ Putting value of π‘=π ππβ‘π₯+πππ π₯ =(βπ)/(π¬π’π§β‘π + πππβ‘π ) +πͺ

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.