Ex 7.10

Chapter 7 Class 12 Integrals
Serial order wise

### Transcript

Ex 7.10, 8 By using the properties of definite integrals, evaluate the integrals : β«_0^(π/4)βlogβ‘(1+tanβ‘π₯ ) ππ₯ Let I=β«_0^(π/4)βlogβ‘γ (1+tanβ‘π₯ )γ ππ₯ β΄ I=β«_0^(π/4)βlogβ‘[1+π­ππ§β‘(π/πβπ) ] ππ₯ I=β«_0^(π/4)βlogβ‘[1+(tanβ‘ π/4 βtanβ‘π₯)/(1 +γ tanγβ‘ π/4 . tanβ‘π₯ )] ππ₯ I=β«_0^(π/4)βlogβ‘[1+(1 β tanβ‘π₯)/(1 + 1 . tanβ‘π₯ )] ππ₯ I=β«_0^(π/4)βlogβ‘[(1 β tanβ‘π₯ + 1 β tanβ‘π₯)/(1 + tanβ‘π₯ )] ππ₯ I=β«_π^(π/π)βπππβ‘[π/(π + πππβ‘π )] ππ Using logβ‘(π/π)=logβ‘πβlogβ‘π I=β«_0^(π/4)β[logβ‘2 βlogβ‘(1+tanβ‘π₯ ) ] ππ₯ π=β«_π^(π/π)βπππβ‘π ππββ«_π^(π/π)βπππβ‘(π+πππβ‘π ) ππ Adding (1) and (2) i.e. (1) + (2) I+I=β«_0^(π/4)βlogβ‘(1+tanβ‘π₯ ) ππ₯+β«_0^(π/4)βlogβ‘2 ππ₯ββ«_0^(π/4)βlogβ‘(1+tanβ‘π₯ ) ππ°=β«_π^(π/π)βπππβ‘π ππ 2I=logβ‘γ 2γ β«_0^(π/4)βππ₯ I=logβ‘γ 2γ/2 [π₯]_0^(π/4) I=logβ‘2/2 [π/4 β 0] I=logβ‘2/2Γπ/4 π°=π/π πππβ‘π