Integration Full Chapter Explained - Integration Class 12 - Everything you need

Last updated at Dec. 20, 2019 by Teachoo
Transcript
Ex 7.9, 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₯ (Using β«1β1/(π₯^2 + π^2 ) ππ₯=1/π tan^(β1)β‘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
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