
Ex 7.8
Last updated at Dec. 16, 2024 by Teachoo
Ex 7.8, 10 β«_0^1βππ₯/(1 + π₯2) Let F(π₯)=β«1βππ₯/(1 + π₯^2 ) =β«1β1/(1^2 + π₯^2 ) ππ₯ =1/1 .tan^(β1)β‘(π₯/1) =tan^(β1) π₯ Hence F(π₯)=tan^(β1) π₯ Now, β«_0^1βγππ₯/(1 + π₯^2 )=πΉ(1)βπΉ(0) γ =tan^(β1)β‘γ(1)βtan^(β1)β‘(0) γ =π/4β0 =π /π