Ex 7.7, 8 - Chapter 7 Class 12 Integrals
Last updated at Dec. 20, 2019 by Teachoo
Transcript
Ex 7.7, 8 β(π₯2+3π₯) β«1βγβ(π₯^2+3π₯) ππ₯γ =β«1βγβ(π₯^2+2(3/2)(π₯) ) ππ₯γ =β«1βγβ(π₯^2+2(3/2)(π₯)+(3/2)^2β(3/2)^2 ) ππ₯γ =β«1βγβ((π₯+3/2)^2β(3/2)^2 ) ππ₯γ It is of the form β«1βγβ(π₯^2βπ^2 ) ππ₯=π₯/2 β(π₯^2βπ^2 )βπ^2/2 πππ|π₯+β(π₯^2βπ^2 )|+πΆγ β΄ Replacing π₯ by π₯+3/2 and a by 3/2 , we get =(π₯ + 3/2)/2 β((π₯+3/2)^2β(3/2)^2 )β|π₯+3/2+β((π₯+π₯/2)^2β(3/2)^2 )|+πΆ =(2π₯ + 3)/4 β((π₯+3/2)^2β9/4) β 9/8 πππ|π₯+3/2 β((π₯+3/2)^2β9/4)|+πΆ =(2π₯ + 3)/4 β(π₯^2+9/4+2(π₯)(3/2)β9/4) β 9/8 πππ|π₯+3/2+β(π₯^2+9/4+2 (π₯)(3/2)β9/4)|+πΆ =(2π₯ + 3)/4 β(π₯^2+2π₯(3/2) )β 9/8 πππ|π₯+3/2+β(π₯^2+2π₯(3/2) )|+πΆ =(ππ + π)/π β(π^π+ππ)β π/π πππ|π+π/π+β(π^π+ππ)|+πͺ
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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.