Last updated at Dec. 16, 2024 by Teachoo
Ex 7.10, 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+πππ§β‘(π /πβπ) ] 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₯ I=β«_0^(π/4)βlogβ‘[(1 β tanβ‘π₯ + 1 β tanβ‘π₯)/(1 + tanβ‘π₯ )] ππ₯ I=β«_π^(π /π)βπππβ‘[π/(π + πππβ‘π )] π π Using logβ‘(π/π)=logβ‘πβlogβ‘π I=β«_0^(π/4)β[logβ‘2 βlogβ‘(1+tanβ‘π₯ ) ] ππ₯ π=β«_π^(π /π)βπππβ‘π π πββ«_π^(π /π)βπππβ‘(π+πππβ‘π ) π π 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₯ ) ππ°=β«_π^(π /π)βπππβ‘π π π 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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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo