Integration Full Chapter Explained - Integration Class 12 - Everything you need
Example 36 - Chapter 7 Class 12 Integrals
Last updated at Dec. 23, 2019 by Teachoo
Transcript
Example 36 Evaluate β«_0^(π/2 )βlogβ‘sinβ‘π₯ ππ₯ Let I1=β«_0^(π/2 )βπππ(π πππ₯) ππ₯ β΄ I1=β«_0^(π/2)βπ ππ(π/2βπ₯)ππ₯ I1= β«_0^(π/2)βπππ(cosβ‘π₯ )ππ₯ Adding (1) and (2) i.e. (1) + (2) I1+ I1=β«_0^(π/2)βγπππ(sinβ‘π₯ )ππ₯+β«_0^(π/2)βπππ(cosβ‘π₯ )ππ₯γ Using Property P4 β«_0^πβγπ(π₯)ππ₯=γ β«_0^πβπ(πβπ₯)ππ₯ β¦(1) β¦(2) 2I1 =β«_0^(π/2)βγlogβ‘[sinβ‘γπ₯ cosβ‘π₯ γ ] ππ₯γ 2I1 = β«_0^(π/2)βγlogβ‘[2sinβ‘γπ₯ cosβ‘π₯ γ/2] ππ₯γ 2I1 = β«_0^(π/2)β[log[2sinβ‘γπ₯ cosβ‘π₯ γ ]βlogβ‘2 ]ππ₯ 2I1 = β«_0^(π/2)β[log[sinβ‘2π₯ ]βlogβ‘2 ]ππ₯ 2I1 = β«_0^(π/2)βlog[sinβ‘2π₯ ]ππ₯ββ«_0^(π/2)βγlog 2 ππ₯γ Solving ππ I2=β«_0^(π/2)βγlog sinβ‘2π₯ ππ₯γ "(Using" logβ‘π + logβ‘π = logβ‘(π.π)) ("Using " log(π/π) = logβ‘(π) β logβ‘(π)) ("Using" sinβ‘2π₯=2 sinβ‘γπ₯ cosβ‘π₯ γ) Let 2π₯=π‘ Differentiating both sides w.r.t.π₯ 2=ππ‘/ππ₯ ππ₯=ππ‘/2 β΄ Putting the values of t and ππ‘ and changing the limits, I2 =β«_0^(π/2)βlog(sinβ‘2π₯ )ππ₯ I2 = β«_0^πβγlog(sinβ‘π‘ ) ππ‘/2γ I2 = 1/2 β«_0^πβlog(sinβ‘π‘ )ππ‘ Using the Property, P6 :- β«_0^2πβπ(π₯)ππ₯=2β«_0^πβπ(π₯)ππ₯, ππ π(2πβπ₯)=π(π₯) Here, π(π‘)=logβ‘π πππ‘ π(2πβπ‘)=π(2πβπ‘)=logβ‘π ππ(2πβπ‘)=logβ‘sinβ‘π‘ Since π(π‘)=π(2πβπ‘) β΄ I2 = 1/2 β«_0^πβlogβ‘sinβ‘γπ‘ ππ‘γ =1/2 Γ2β«_0^(π/2)βlogβ‘sinβ‘γπ‘. ππ‘γ =β«_0^(π/2)βlogβ‘sinβ‘γπ‘. ππ‘γ I2=β«_0^(π/2)βlogβ‘sinβ‘γπ₯ ππ₯γ Now, Using the Property, P0 :- β«_π^πβπ(π₯)ππ₯=β«_π^πβπ(π‘)ππ‘ Putting the value of I2 in equation (3), we get 2I1 =β«_π^(π /π)βπ₯π¨π [πππβ‘ππ ]β‘π π ββ«_0^(π/2)βlog(2)β‘ππ₯ 2I1 = β«_π^(π /π)βπ₯π¨π (πππβ‘π )β‘π π βlog(2) β«_0^(π/2)βγ1.γβ‘ππ₯ 2I1 = ππ β log(2) [π₯]_0^(π/2) 2I1βI1=βlogβ‘2 [π/2β0] I1=βlogβ‘2 [π/2] β΄ ππ=(β π )/π π₯π¨π β‘π
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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.