Ex 7.3, 16 - Chapter 7 Class 12 Integrals
Last updated at April 16, 2024 by Teachoo
Ex 7.3
Ex 7.3, 1
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Ex 7.3, 16 β«1βγtan^4 π₯γ ππ₯ β«1βγtan^4 π₯γ ππ₯=β«1βγtan^2 π₯ .tan^2 π₯γ ππ₯ =β«1βγ(sec^2β‘π₯β 1) tan^2β‘π₯ γ ππ₯ =β«1β(sec^2β‘π₯.tan^2β‘π₯βtan^2β‘π₯ ) ππ₯ =β«1βγtan^2β‘π₯.sec^2β‘π₯ γ ππ₯ββ«1βγtan^2 π₯γ ππ₯ Solving both these integrals separately We know that γπ‘ππγ^2 π=γπ ππγ^2β‘γπβ1γ β«1βγγπππγ^πβ‘π.γπππγ^πβ‘π γ π π Let tan π₯=π‘ sec^2β‘π₯=ππ‘/ππ₯ ππ₯=1/sec^2β‘π₯ . ππ‘ Now, β«1βtan^2β‘π₯ .sec^2β‘π₯.ππ₯ =β«1βπ‘^2 .sec^2β‘π₯. 1/sec^2β‘π₯ .ππ‘ =β«1βπ‘^2 . ππ‘ =π‘^(2 + 1)/(2 + 1) + C =π‘^3/3+πΆ Putting value of π‘=π‘ππβ‘π₯ =tan^3β‘π₯/3+πΆ1 β«1βγγπππγ^π πγ π π =β«1β(sec^2β‘π₯β1) ππ₯ =β«1βsec^2β‘π₯ ππ₯ββ«1β1β‘γ.ππ₯γ =tanβ‘π₯βπ₯+πΆ2 "As" β«1βγπ₯^π ππ₯=π₯^(π+1)/(π+1)+πΆγ & β«1βsec^2β‘π₯ ππ₯=tanβ‘π₯+πΆ Now, β«1βγtan^4 π₯γ ππ₯=β«1βγtan^2 π₯ .sec^2 π₯γ ππ₯ββ«1βγtan^2 π₯γ ππ₯ =tan^3β‘π₯/3+πΆ1β(tanβ‘π₯βπ₯+πΆ2) =tan^3β‘π₯/3 βtanβ‘π₯+π₯+πΆ1βπΆ2 =γπππγ^πβ‘π/π βπππβ‘π+π+πͺ (Where πΆ=πΆ1βπΆ2)
