Ex 7.10

Chapter 7 Class 12 Integrals
Serial order wise

### Transcript

Ex 7.10, 16 By using the properties of definite integrals, evaluate the integrals : β«_0^πβlogβ‘(1+cosβ‘π₯ ) ππ₯ Let I=β«_π^πβπππβ‘(π+πππβ‘π ) ππ β΄ I=β«_0^πβlogβ‘(1+πππ (πβπ₯)) ππ₯ π=β«_π^πβπ₯π¨π β‘(πβπππβ‘π ) ππ₯ Adding (1) and (2) i.e. (1) + (2) I+I=β«_0^πβπππ(1+cosβ‘π₯ )ππ₯+β«_0^πβπππ(1βcosβ‘π₯ )ππ₯ ππ=β«_π^πβ[πππ(π+πππβ‘π )+πππ(πβπππβ‘π )]ππ "Using log(a) + log(b)" = "log(a.b)") 2I=β«_0^πβπππ[(π+πππβ‘π )(πβπππβ‘π )]ππ 2I=β«_0^πβπππ[1βcos^2β‘π₯ ]ππ₯ 2I=β«_0^πβπππ(γπ ππγ^2β‘π₯ )ππ₯ Using log π^π=π πππβ‘π 2I=β«_0^πβγ2 πππ(π ππβ‘π₯ )ππ₯γ 2I=2β«_0^πβπππ(π ππβ‘π₯ )ππ₯ π=β«_π^πβπππ(πππβ‘π )ππ Here, π(π)=logβ‘sinβ‘π₯ f(πβπ)=πππ[π ππ(πβπ₯)]ππ₯ =πππ(sinβ‘π₯ )ππ₯ =π(π) Therefore, I=β«_π^πβπππ(π¬π’π§β‘π )ππ=πβ«_π^(π/π)βπππ(πππβ‘π )ππ Let I1=β«_0^(π/2 )βπππ(π πππ₯) ππ₯ Solving ππ I1=β«_0^(π/2 )βπππ(π πππ₯) ππ₯ β΄ I1=β«_π^(π/π)βπππ(π/πβπ)ππ I1= β«_π^(π/π)βπππ(πππβ‘π )ππ Adding (2) and (3) i.e. (2) + (3) ππ + ππ =β«_π^(π/π)βγπππ(πππβ‘π )ππ+β«_π^(π/π)βπππ(ππ¨π¬β‘π )ππγ "Using" πππβ‘π + πππβ‘π = πππβ‘(π.π) 2I1 =β«_π^(π/π)βγπ₯π¨π β‘[π¬π’π§β‘γπ ππ¨π¬β‘π γ ] ππγ 2I1 = β«_π^(π/π)βγπππβ‘[ππππβ‘γπ πππβ‘π γ/π] ππγ "Using " πππ(π/π) = logβ‘(π) β logβ‘(π)) 2I1 = β«_0^(π/2)β[log[2sinβ‘γπ₯ cosβ‘π₯ γ ]βπππ(2)]ππ₯ "Using" πππβ‘ππ=2 sinβ‘γπ₯ cosβ‘π₯ γ 2I1 = β«_0^(π/2)β[log[sinβ‘2π₯ ]βπππ2]ππ₯ 2I1 = β«_π^(π/π)βπ₯π¨π [πππβ‘ππ ]ππββ«_0^(π/2)βlog(2)ππ₯ Solving ππ ππ=β«_π^(π/π)βγπ₯π¨π  πππβ‘ππ 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 ππ = π/π β«_π^πβπ₯π¨π (πππβ‘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‘ As, π(π)=π(ππβπ) β΄ ππ = 1/2 β«_0^πβlogβ‘sinβ‘γπ‘ ππ‘γ =π/π Γ πβ«_π^(π/π)βπππβ‘πππβ‘γπ. ππγ =β«_0^(π/2)βlogβ‘sinβ‘γπ‘. ππ‘γ =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«_π^(π/π)βγπ.γβ‘ππ 2I1 = I1 β log(2) [π₯]_0^(π/2) 2I1βI1=βlogβ‘2 [π/2β0] ππ=βπππβ‘π [π/π] Hence, π=2 I1= 2 Γ (βπ)/2 logβ‘2 =βπ πππβ‘π