Integration using trigo identities - sin^2,cos^2 etc formulae

Chapter 7 Class 12 Integrals
Concept wise

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Ex 7.3, 19 Integrate the function 1/(sinβ‘π₯ . cos^3β‘π₯ ) β«1β1/(sinβ‘π₯ . cos^3β‘π₯ ) ππ₯ =β«1β(sin^2β‘π₯ + cos^2β‘π₯)/(sinβ‘π₯ . cos^3β‘π₯ ) ππ₯ =β«1β(sin^2β‘π₯/(sinβ‘π₯ . cos^3β‘π₯ )+cos^2β‘π₯/(sinβ‘π₯ . cos^3β‘π₯ )) ππ₯ =β«1β(π ππβ‘π₯/cos^3β‘π₯ +πππ β‘π₯/(sinβ‘π₯ . cos^2β‘π₯ )) ππ₯ =β«1β(π ππβ‘π₯/πππ β‘π₯ Γ1/cos^2β‘π₯ +πππ β‘π₯/sinβ‘π₯ Γ1/cos^2β‘π₯ ) ππ₯ =β«1βγ1/cos^2β‘π₯ (π ππβ‘π₯/πππ β‘π₯ +πππ β‘π₯/sinβ‘π₯ ) γ ππ₯ (As γπ ππγ^2β‘π+γπππ γ^2β‘π=1) =β«1βγsec^2β‘π₯ (tanβ‘π₯+cotβ‘π₯ ) γ ππ₯ =β«1βγsec^2β‘π₯ (tanβ‘π₯+1/tanβ‘π₯ ) γ ππ₯ =β«1βγ(tanβ‘π₯+1/tanβ‘π₯ ). sec^2β‘π₯ γ ππ₯ Putting π‘ππβ‘π₯=π‘ Differentiating w.r.t.x sec^2β‘π₯=ππ‘/ππ₯ ππ₯=1/sec^2β‘π₯ ππ‘ Thus, our equation becomes =β«1βγ(π‘+1/π‘) sec^2β‘π₯Γ1/sec^2β‘π₯ γ ππ‘ =β«1β(π‘+1/π‘) ππ‘ =β«1βπ‘ ππ‘+β«1β1/π‘ ππ‘ =π‘^2/2 +logβ‘|π‘|+πΆ Putting value of π‘=π‘ππβ‘π₯ =(γπ‘ππγ^2 π₯)/2+πππβ‘|π‘ππβ‘π₯ |+πΆ =πππβ‘|πππβ‘π |+(γπππγ^π π)/π +πͺ