Last updated at Dec. 16, 2024 by Teachoo
Ex 7.10, 1 By using the properties of definite integrals, evaluate the integrals : β«_0^(π/2)βγcos^2β‘π₯ ππ₯γ Let π=β«_π^(π /π)βγγπππγ^πβ‘π π πγ I=β«_π^(π /π)βγγππ¨π¬γ^πβ‘ (π /πβπ)π πγ I= β«_π^((π )/π)βγγπ¬π’π§γ^π πγβ‘π π Adding (1) and (2) I+I= β«_0^(π/2)βγcos^2β‘π₯ ππ₯γ + β«_0^((π )/2)βγsin^2 π₯γβ‘ππ₯ 2I= β«_0^((π )/2)β(cos^2β‘γπ₯+sin^2β‘π₯ γ )β‘ππ₯ ππ =β«_π^((π )/π)βγπ .γβ‘π π 2I=[π₯]_0^(π/2) 2I =π/2β0 2I =π/2 π=π /π Evaluate: β«_0^πβγπ^cosβ‘π₯ /(π^cosβ‘π₯ + π^γβcosγβ‘π₯ ) ππ₯γ Let I=β«_0^πβγπ^cosβ‘π₯ /(π^cosβ‘π₯ + π^γβcosγβ‘π₯ ) ππ₯γ " " I= β«_0^πβγπ^cosβ‘γ(π β π₯)γ /(π^cosβ‘γ(π β π₯)γ + π^γβcosγβ‘γ(π β π₯)γ ) ππ₯γ " " I=β«_0^πβγπ^γβcosγβ‘π₯ /(π^γβcosγβ‘π₯ + π^(γβ(βcosγβ‘π₯)) ) ππ₯γ I=β«_0^πβγπ^γβcosγβ‘π₯ /(π^γβcosγβ‘π₯ + π^cosβ‘π₯ ) ππ₯γ Evaluate: β«_0^πβγπ^cosβ‘π₯ /(π^cosβ‘π₯ + π^γβcosγβ‘π₯ ) ππ₯γ Let I=β«_0^πβγπ^cosβ‘π₯ /(π^cosβ‘π₯ + π^γβcosγβ‘π₯ ) ππ₯γ " " I= β«_0^πβγπ^cosβ‘γ(π β π₯)γ /(π^cosβ‘γ(π β π₯)γ + π^γβcosγβ‘γ(π β π₯)γ ) ππ₯γ " " I=β«_0^πβγπ^γβcosγβ‘π₯ /(π^γβcosγβ‘π₯ + π^(γβ(βcosγβ‘π₯)) ) ππ₯γ I=β«_0^πβγπ^γβcosγβ‘π₯ /(π^γβcosγβ‘π₯ + π^cosβ‘π₯ ) ππ₯γ Adding (1) and (2) i.e. (1) + (2) I+I=β«_0^πβγπ^cosβ‘π₯ /(π^cosβ‘π₯ + π^γβcosγβ‘π₯ ) ππ₯γ + β«_0^πβγπ^γβcosγβ‘π₯ /(π^γβcosγβ‘π₯ + π^cosβ‘π₯ ) ππ₯γ 2I=β«_0^πβγ(π^cosβ‘π₯ + π^γβcosγβ‘π₯ )/(π^cosβ‘π₯ + π^γβcosγβ‘π₯ ) ππ₯γ 2I =β«_0^πβγ1 .γβ‘ππ₯ 2I=[π₯]_0^π 2I =πβ0 2I =π π=π /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