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Ex 7.3
Last updated at April 16, 2024 by Teachoo
Ex 7.3, 14 Integrate the function 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/(π + 2 sinβ‘π₯ cosβ‘π₯ ) ππ₯ =β«1βcosβ‘γπ₯ βγ sinγβ‘π₯ γ/(γπ¬π’π§γ^πβ‘π + γππ¨π¬γ^πβ‘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 ππβ‘2 π=2 π ππβ‘π πππ β‘π) (As γπ ππγ^2β‘π+γπππ γ^2β‘π=1) 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