Integration Full Chapter Explained - Integration Class 12 - Everything you need


Last updated at Dec. 20, 2019 by Teachoo
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Ex 7.11, 8 By using the properties of definite integrals, evaluate the integrals : β«_0^(π/4)βlogβ‘(1+tanβ‘π₯ ) ππ₯ Let I=β«_0^(π/4)βlogβ‘γ (1+tanβ‘π₯ )γ ππ₯ β΄ I=β«_0^(π/4)βlogβ‘[1+tanβ‘(π/4βπ₯) ] ππ₯ I=β«_0^(π/4)βlogβ‘[1+(tanβ‘ π/4 βtanβ‘π₯)/(1 +γ tanγβ‘ π/4 . tanβ‘π₯ )] ππ₯ I=β«_0^(π/4)βlogβ‘[1+(1 β tanβ‘π₯)/(1 + 1 . tanβ‘π₯ )] ππ₯ Ex 7.11, 8 By using the properties of definite integrals, evaluate the integrals : β«_0^(π/4)βlogβ‘(1+tanβ‘π₯ ) ππ₯ Let I=β«_0^(π/4)βlogβ‘γ (1+tanβ‘π₯ )γ ππ₯ β΄ I=β«_0^(π/4)βlogβ‘[1+tanβ‘(π/4βπ₯) ] ππ₯ I=β«_0^(π/4)βlogβ‘[1+(tanβ‘ π/4 βtanβ‘π₯)/(1 +γ tanγβ‘ π/4 . tanβ‘π₯ )] ππ₯ I=β«_0^(π/4)βlogβ‘[1+(1 β tanβ‘π₯)/(1 + 1 . tanβ‘π₯ )] ππ₯ (tan(πβπ)=tanβ‘γπ β π‘ππ πγ/(1+γ tanγβ‘γπ tanβ‘π γ )) (As tan(π/4)=1) I=β«_0^(π/4)βlogβ‘[(1 β tanβ‘π₯ + 1 β tanβ‘π₯)/(1 + tanβ‘π₯ )] ππ₯ I=β«_0^(π/4)βlogβ‘[2/(1 + tanβ‘π₯ )] ππ₯ I=β«_0^(π/4)β[logβ‘2 βlogβ‘(1+tanβ‘π₯ ) ] ππ₯ I=β«_0^(π/4)βlogβ‘2 ππ₯ββ«_0^(π/4)βlogβ‘(1+tanβ‘π₯ ) ππ₯ Adding (1) and (2) i.e. (1) + (2) I+I=β«_0^(π/4)βlogβ‘(1+tanβ‘π₯ ) ππ₯+β«_0^(π/4)βlogβ‘2 ππ₯ββ«_0^(π/4)βlogβ‘(1+tanβ‘π₯ ) 2I=β«_0^(π/4)βlogβ‘2 ππ₯ (Using logβ‘(π/π) =logβ‘πβlogβ‘π) β¦(2) 2I=logβ‘γ 2γ β«_0^(π/4)βππ₯ I=logβ‘γ 2γ/2 [π₯]_0^(π/4) I=logβ‘2/2 [π/4 β 0] I=logβ‘2/2Γπ/4 π°=π /π πππβ‘π
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