Examples

Chapter 7 Class 12 Integrals
Serial order wise

### Transcript

Example 34 Evaluate β«_0^(π/2 )βlogβ‘sinβ‘π₯ ππ₯ Let I1=β«_0^(π/2 )βπππ(π πππ₯) ππ₯ β΄ I1=β«_0^(π/2)βπ ππ(π/2βπ₯)ππ₯ I1= β«_0^(π/2)βπππ(cosβ‘π₯ )ππ₯ Adding (1) and (2) i.e. (1) + (2) I1+ I1=β«_0^(π/2)βγπππ(sinβ‘π₯ )ππ₯+β«_0^(π/2)βπππ(cosβ‘π₯ )ππ₯γ 2I1 =β«_0^(π/2)βγlogβ‘[sinβ‘γπ₯ cosβ‘π₯ γ ] ππ₯γ 2I1 = β«_0^(π/2)βγlogβ‘[2sinβ‘γπ₯ cosβ‘π₯ γ/2] ππ₯γ 2I1 = β«_0^(π/2)β[log[2sinβ‘γπ₯ cosβ‘π₯ γ ]βlogβ‘2 ]ππ₯ 2I1 = β«_0^(π/2)β[log[sinβ‘2π₯ ]βlogβ‘2 ]ππ₯ 2I1 = β«_0^(π/2)βlog[sinβ‘2π₯ ]ππ₯ββ«_0^(π/2)βγlog 2 ππ₯γ Solving ππ I2=β«_0^(π/2)βγlog sinβ‘2π₯ ππ₯γ Let 2π₯=π‘ Differentiating both sides w.r.t.π₯ 2=ππ‘/ππ₯ ππ₯=ππ‘/2 β΄ Putting the values of t and ππ‘ and changing the limits, I2 =β«_0^(π/2)βlog(sinβ‘2π₯ )ππ₯ I2 = β«_0^πβγlog(sinβ‘π‘ ) ππ‘/2γ I2 = 1/2 β«_0^πβlog(sinβ‘π‘ )ππ‘ Here, π(π‘)=logβ‘π πππ‘ π(2πβπ‘)=π(2πβπ‘)=logβ‘π ππ(2πβπ‘)=logβ‘sinβ‘π‘ Since π(π‘)=π(2πβπ‘) β΄ I2 = 1/2 β«_0^πβlogβ‘sinβ‘γπ‘ ππ‘γ =1/2 Γ2β«_0^(π/2)βlogβ‘sinβ‘γπ‘. ππ‘γ =β«_0^(π/2)βlogβ‘sinβ‘γπ‘. ππ‘γ I2=β«_0^(π/2)βlogβ‘sinβ‘γπ₯ ππ₯γ Putting the value of I2 in equation (3), we get 2I1 =β«_π^(π/π)βπ₯π¨π [πππβ‘ππ ]β‘ππ ββ«_0^(π/2)βlog(2)β‘ππ₯ 2I1 = β«_π^(π/π)βπ₯π¨π (πππβ‘π )β‘ππ βlog(2) β«_0^(π/2)βγ1.γβ‘ππ₯ 2I1 = ππ β log(2) [π₯]_0^(π/2) 2I1βI1=βlogβ‘2 [π/2β0] I1=βlogβ‘2 [π/2] β΄ ππ=(β π)/π π₯π¨π β‘π

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#### Davneet Singh

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, Social Science, Physics, Chemistry, Computer Science at Teachoo.