Integration using trigo identities - 3x formulae

Chapter 7 Class 12 Integrals
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Misc 18 Integrate the function 1/β(sin^3β‘π₯ sinβ‘(π₯ + πΌ) ) Solving sin^3β‘π₯ sinβ‘(π₯ + πΌ) =sin^3β‘π₯ [sinβ‘π₯ cosβ‘πΌ+cosβ‘π₯.sinβ‘πΌ ] =γsin^4 π₯γβ‘cosβ‘πΌ +cosβ‘π₯.sin^3β‘π₯ sinβ‘πΌ =γsin^4 π₯γβ‘cosβ‘πΌ +cosβ‘π₯.sin^3β‘π₯ sinβ‘πΌΓsinβ‘π₯/sinβ‘π₯ =sin^4β‘π₯ [cosβ‘πΌ+cosβ‘π₯ . sinβ‘πΌ.1/sinβ‘π₯ ] Hence Using π ππβ‘(π΄+π΅)=π ππβ‘π΄ πππ β‘π΅+πππ β‘π΄.π ππβ‘π΅ =sin^4β‘π₯ [cosβ‘πΌ+cosβ‘π₯/sinβ‘π₯ . sinβ‘πΌ ] =sin^4β‘π₯ [cosβ‘πΌ+cotβ‘π₯ sinβ‘πΌ ] Therefore sin^3β‘π₯ sinβ‘(π₯+πΌ)=sin^4β‘π₯ (cosβ‘πΌ+cotβ‘π₯.sinβ‘πΌ ) Now β«1β1/β(sin^3β‘π₯ sinβ‘(π₯ + πΌ) ) ππ₯ =β«1β1/β(sin^4β‘π₯ (cosβ‘πΌ + cotβ‘π₯ . sinβ‘πΌ ) ) ππ₯ =β«1βγ1/β(sin^4β‘π₯ )Γ1/β(cosβ‘πΌ + cotβ‘π₯ . sinβ‘πΌ )γ ππ₯ =β«1βγ1/sin^2β‘π₯ Γ1/β(cosβ‘πΌ + cotβ‘π₯ . sinβ‘πΌ )γ ππ₯ Let cosβ‘πΌ+cotβ‘π₯. sinβ‘πΌ=π‘ Diff w.r.t. x π(cosβ‘πΌ + cotβ‘π₯ sinβ‘πΌ )/ππ₯=ππ‘/ππ₯ π(cosβ‘πΌ )/ππ₯+sinβ‘πΌ π(cotβ‘π₯ )/ππ₯=ππ‘/ππ₯ =β«1βγ1/β(sin^4β‘π₯ )Γ1/β(cosβ‘πΌ + cotβ‘π₯ . sinβ‘πΌ )γ ππ₯ =β«1βγ1/sin^2β‘π₯ Γ1/β(cosβ‘πΌ + cotβ‘π₯ . sinβ‘πΌ )γ ππ₯ Let cosβ‘πΌ+cotβ‘π₯. sinβ‘πΌ=π‘ Diff w.r.t. x π(cosβ‘πΌ + cotβ‘π₯ sinβ‘πΌ )/ππ₯=ππ‘/ππ₯ π(cosβ‘πΌ )/ππ₯+sinβ‘πΌ π(cotβ‘π₯ )/ππ₯=ππ‘/ππ₯ 0+sinβ‘πΌ (βπππ ππ^2 π₯)=ππ‘/ππ₯ βsinβ‘πΌ πππ ππ^2 π₯=ππ‘/ππ₯ ππ₯=ππ‘/(βsinβ‘πΌ πππ ππ^2 π₯) ππ₯=1/(βsinβ‘πΌ ) . 1/(πππ ππ^2 π₯) . ππ‘ ππ₯=1/(βsinβ‘πΌ ) . sin^2β‘π₯. ππ‘ Now our equation becomes β«1β1/(sin^2β‘π₯ β(cosβ‘πΌ + cotβ‘π₯ sinβ‘πΌ ) ) ππ₯ =β«1β1/(sin^2β‘π₯ βπ‘ )Γ1/(βsinβ‘πΌ )Γsin^2β‘π₯ ππ‘ cosβ‘πΌ &sinβ‘πΌ "are constant" β(" " @"&" π(cotβ‘π₯ )/ππ₯=βπππ ππ π₯) =1/(βsinβ‘πΌ ) β«1β1/βπ‘ ππ‘ =(β1)/sinβ‘πΌ β«1β(π‘)^((β1)/2) ππ‘ =(β1)/sinβ‘πΌ [π‘^((β1)/2 + 1)/((β1)/2 + 1) +πΆ] =(β1)/sinβ‘πΌ [π‘^(1/2)/(1/2) +πΆ] =(βπ)/πππβ‘πΆ [πβπ +πͺ] Putting back value of π‘=β(πππ β‘πΌ+πππ‘β‘π₯. π ππβ‘πΌ ) =(β1)/sinβ‘πΌ [2β(πππ β‘πΌ+πππ‘β‘π₯. π ππβ‘πΌ )+πΆ] =(β2)/sinβ‘πΌ β(πππ β‘πΌ+πππ‘β‘π₯. π ππβ‘πΌ ) β1/sinβ‘πΌ . πΆ =(β2)/sinβ‘πΌ β(πππ β‘πΌ+πππ‘β‘π₯. π ππβ‘πΌ ) +πΆ Now, From (1) sin^3β‘π₯ sinβ‘(π₯+πΌ)=sin^4β‘π₯ (πππ β‘πΌ+πππ‘β‘π₯. π ππβ‘πΌ ) (sin^3β‘π₯ sinβ‘(π₯ + πΌ))/sin^4β‘π₯ =πππ β‘πΌ+πππ‘β‘π₯. π ππβ‘πΌ γπ ππ γβ‘(π₯ + πΌ)/sinβ‘π₯ =πππ β‘πΌ+πππ‘β‘π₯. π ππβ‘πΌ Thus, Answer =(βπ)/πππβ‘π β(πππβ‘(π + πΆ)/π¬π’π§β‘π ) + πͺ