Ex 7.10

Chapter 7 Class 12 Integrals
Serial order wise

### Transcript

Ex 7.10, 10 By using the properties of definite integrals, evaluate the integrals : β«_0^(π/2)βγ (2 logβ‘sinβ‘π₯ βlogβ‘sinβ‘2π₯ ) γ ππ₯ Let I1=β«_0^(π/2)βγ (2 logβ‘sinβ‘π₯ βlogβ‘sinβ‘2π₯ ) γ ππ₯ I1= β«_0^(π/2)βγ [2 logβ‘sinβ‘π₯ βπππ(2 sinβ‘γπ₯ cosβ‘π₯ γ )] γ ππ₯ I1= β«_0^(π/2)βγ [2 logβ‘sinβ‘π₯ βlogβ‘2βlogβ‘sinβ‘γπ₯βlogβ‘cosβ‘π₯ γ ] γ ππ₯ I1= β«_0^(π/2)βγ [logβ‘sinβ‘π₯ βπππ2βlogβ‘cosβ‘π₯ ] γ ππ₯ I1= β«_0^(π/2)βlogβ‘sinβ‘γπ₯ ππ₯γ ββ«_0^(π/2)βγlogβ‘2ππ₯ββ«_0^(π/2)βlogβ‘cosβ‘γπ₯ ππ₯γ γ Solving I2 I2=β«_0^(π/2)βlogβ‘cosβ‘γπ₯ ππ₯γ β΄ I2= β«_0^(π/2)βlogβ‘πππ (π/2βπ₯)ππ₯ I2=β«_0^(π/2)βlogβ‘sinβ‘γπ₯ ππ₯γ Put the value of I2 in (1) i.e. I1 β΄ I1= β«_0^(π/2)βlogβ‘sinβ‘γπ₯ ππ₯γ ββ«_0^(π/2)βγlog 2γβ‘ππ₯ ββ«_π^(π/π)βπ₯π¨π β‘ππ¨π¬β‘γπ ππγ I1= β«_0^(π/2)βlogβ‘sinβ‘γπ₯ ππ₯γ ββ«_0^(π/2)βγlog 2γβ‘ππ₯ ββ«_π^(π/π)βπππβ‘π¬π’π§β‘γπ ππγ I1= β β«_0^(π/2)βγlog 2γβ‘ππ₯ I1= β log 2β«_0^(π/2)βππ₯ I1= β log 2[π₯]_0^(π/2) I1= β log 2[π/2β0] I1= β log 2Γπ/2 I1= logβ‘γγ (2)γ^(β1) γ [π/2] I1= π/π π₯π¨π  (π/π)