Ex 7.3, 9 - Chapter 7 Class 12 Integrals
Last updated at April 16, 2024 by Teachoo
Ex 7.3
Ex 7.3, 1
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Transcript
Ex 7.3, 9 Integrate (πππ π₯)/(1 + πππ π₯) β«1βγ(πππ π₯)/(1 + πππ π₯) " " ππ₯γ = β«1β((cosβ‘π₯ + 1 β 1)/(1 + cosβ‘π₯ )) ππ₯ =β«1β((1 + cosβ‘π₯ β 1)/(1 + cosβ‘π₯ )) ππ₯ =β«1β((1 + cosβ‘π₯)/(1 + cosβ‘π₯ ) β 1/(1 + cosβ‘π₯ )) ππ₯ =β«1βγ1β1/(1 + cosβ‘π₯ )γ ππ₯ =β«1β1 ππ₯ββ«1βπ/(π + πππβ‘π ) ππ₯ =β«1β1 ππ₯ββ«1β1/(π γπππγ^πβ‘γπ/πγ ) ππ₯ =β«1β1 ππ₯ββ«1β1/2 sec^2β‘γπ₯/2γ ππ₯ =β«1β1 ππ₯β1/2 β«1βsec^2β‘γπ₯/2γ ππ₯ =π₯β 1/2 γtan γβ‘γπ₯/2γ/(1/2) +πΆ =π₯β 2/2 γtan γβ‘γπ₯/2γ +πΆ =πβ γπππ§ γβ‘γπ/πγ +πͺ β«1βsec^2β‘(ππ₯+π) ππ₯=π‘ππβ‘(ππ₯ + π)/π +πΆ We know that cos 2π=2 cos^2β‘γπβ1γ cosβ‘2π+1=2 cos^2β‘π Replacing π by π₯/2 cosβ‘2(π₯/2)+1=2 cos^2β‘γπ₯/2γ cosβ‘π₯+1=2 cos^2β‘γπ₯/2γ
