Ex 7.2

Chapter 7 Class 12 Integrals
Serial order wise

Transcript

Ex 7.2, 33 Integrate 1/(1 β π‘ππβ‘π₯ ) β«1β1/(1 β tanβ‘π₯ ) ππ₯ = β«1β1/(1 β sinβ‘π₯/cosβ‘π₯ ) ππ₯ = β«1β1/(γcosβ‘π₯ β sinγβ‘π₯/cosβ‘π₯ ) ππ₯ = β«1βcosβ‘π₯/γcosβ‘π₯ β sinγβ‘π₯ ππ₯ = β«1β(2 cosβ‘π₯)/2(γcosβ‘π₯ β sinγβ‘π₯ ) ππ₯ = β«1β(cosβ‘π₯ + cosβ‘π₯)/2(γcosβ‘π₯ β sinγβ‘π₯ ) ππ₯ = β«1β(cosβ‘π₯ + cosβ‘π₯ + sinβ‘π₯ β sinβ‘π₯)/2(γcosβ‘π₯ β sinγβ‘π₯ ) ππ₯ = β«1β(cosβ‘π₯ β sinβ‘π₯ + cosβ‘π₯ + sinβ‘π₯)/2(γcosβ‘π₯ β sinγβ‘π₯ ) ππ₯ = 1/2 β«1β((cosβ‘π₯ β sinβ‘π₯ + cosβ‘π₯ + sinβ‘π₯)/γcosβ‘π₯ β sinγβ‘π₯ ) ππ₯ = 1/2 β«1β((cosβ‘π₯ + sinβ‘π₯)/γcosβ‘π₯ β sinγβ‘π₯ +(cosβ‘π₯ β sinβ‘π₯)/γcosβ‘π₯ β sinγβ‘π₯ ) ππ₯ = 1/2 β«1β(1+ (cosβ‘π₯ + sinβ‘π₯)/γcosβ‘π₯ β sinγβ‘π₯ ) ππ₯ = 1/2 β«1βππ₯ + 1/2 β«1β(cosβ‘π₯ + sinβ‘π₯)/γcosβ‘π₯ β sinγβ‘π₯ ππ₯ (Adding & subtracting π ππβ‘π₯ in numerator) = π₯/2+1/2 β«1β(cosβ‘π₯ + sinβ‘π₯)/γcosβ‘π₯ β sinγβ‘π₯ ππ₯+πΆ_1 Solving π1 I1 = β«1β(cosβ‘π₯ + sinβ‘π₯)/γcosβ‘π₯ β sinγβ‘π₯ .ππ₯ Let cosβ‘π₯ β sinβ‘π₯=π‘ Differentiating both sides π€.π.π‘.π₯ γγ βπ ππγβ‘π₯βπππ γβ‘π₯=ππ‘/ππ₯ β(γπ ππβ‘π₯+πππ γβ‘π₯ )=ππ‘/ππ₯ ππ₯=ππ‘/(β(γsinβ‘π₯ β cosγβ‘π₯ ) ) Thus, our equation becomes I1 = β«1β(cosβ‘π₯ + sinβ‘π₯)/γcosβ‘π₯ β sinγβ‘π₯ . ππ₯ I1 = β«1β(cosβ‘π₯ β sinβ‘π₯)/π‘ . ππ‘/(β(γsinβ‘π₯ β cosγβ‘π₯ ) ) I1 = β1β«1βππ‘/π‘ I1 = βlogβ‘|π‘|+πΆ2 I1 = βlogβ‘|cosβ‘π₯βsinβ‘π₯ |+πΆ2 (Using β«1β1/π₯. ππ₯=πππβ‘|π₯|) (Using γγπ‘=πππ γβ‘π₯βπ ππγβ‘π₯ ) Putting the value of I1 in (1) β΄ β«1βγ1/(1 + tanβ‘π₯ ) " " γ= π₯/2+1/2 β«1β(cosβ‘π₯ + sinβ‘π₯)/γcosβ‘π₯ β sinγβ‘π₯ ππ₯+πΆ_1 = π₯/2+1/2(βπ₯π¨π β‘γ |πππβ‘πβπππβ‘π |γ)+πΆ2/2+πΆ1 = π/π βπ/π πππβ‘γ |πππβ‘πβπππβ‘π |γ+πͺ