Example 22 - Chapter 7 Class 12 Integrals (Term 2)
Last updated at Dec. 20, 2019 by Teachoo
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Example 22 Important You are here
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Example 25 Important Deleted for CBSE Board 2023 Exams
Example 26 Deleted for CBSE Board 2023 Exams
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Example 25 (Supplementary NCERT) Important Deleted for CBSE Board 2023 Exams
Chapter 7 Class 12 Integrals
Serial order wise
Transcript
Example 22 Find (i) β«1βπ^π₯ (tan^(β1)β‘π₯+ 1/(1 + π₯^2 )) ππ₯ β«1βγπ^π₯ (tan^(β1)β‘π₯+1/(1 + π₯^2 ))ππ₯γ It is of the form β«1βγπ^π₯ [π(π₯)+π^β² (π₯)] γ ππ₯=π^π₯ π(π₯)+πΆ Where π(π₯)=tan^(β1)β‘π₯ π^β² (π₯)= 1/(1 + π₯^2 ) So, our equation becomes β«1βγπ^π₯ (tan^(β1)β‘π₯+1/(1 + π₯^2 ))ππ₯γ=π^π γπππ§γ^(βπ)β‘γπ+πͺγ Example 22 Find (ii) β«1β((π₯^2 + 1) π^π₯)/(π₯ + 1)^2 ππ₯ β«1βγ(π₯^2 + 1)/(π₯ + 1)^2 .π^π₯ ππ₯γ Adding and subtracting 1 in numerator =β«1βγ(π₯^2+ 1 + 1 β 1)/(π₯ + 1)^2 .π^π₯ .ππ₯γ =β«1βγ(π₯^2 β 1 + 1 + 1)/(π₯ + 1)^2 .π^π₯ .ππ₯γ =β«1βγ[(π₯^2 β 1)/(π₯ + 1)^2 +2/(π₯ + 1)^2 ] π^π₯ ππ₯γ =β«1βγπ^π₯ [(π₯^2 β (1)^2)/(π₯ + 1)^2 +2/(π₯ + 1)^2 ]ππ₯γ =β«1βγπ^π₯ [(π₯ β 1)(π₯ + 1)/(π₯ + 1)^2 +2/(π₯ + 1)^2 ]ππ₯γ =β«1βγπ^π₯ [(π₯ β 1)/(π₯ + 1)+2/(π₯ + 1)^2 ]ππ₯γ It is of form β«1βγπ^π₯ [π(π₯)+π^β² (π₯)] γ ππ₯=π^π₯ π(π₯)+πΆ Where π(π₯)=(π₯ β 1)/(π₯ + 1) π^β² (π₯)=π/ππ₯ [(π₯ β 1)/(π₯ + 1)] π^β² (π₯)=(1.(π₯ + 1) β1 (π₯ β 1))/(π₯ + 1)^2 =(π₯ + 1 β π₯ + 1)/(π₯ + 1)^2 =2/(π₯ + 1)^2 Thus, our equation becomes β«1βγ(π₯^2 + 1)/(π₯ + 1)^2 .π^π₯=β«1βγπ^π₯ [(π₯ β 1)/(π₯ + 1)+2/(π₯ + 1)^2 ]ππ₯γγ =π^π₯ [(π₯ β 1)/(π₯ + 1)]+πΆ =(π β π)/(π + π).π^π+πͺ
