Example 27 - Chapter 7 Class 12 Integrals
Last updated at April 16, 2024 by Teachoo
Definite Integration - By Substitution
Definite Integration - By Substitution
Last updated at April 16, 2024 by Teachoo
Example 27 (Method 1) Evaluate β«_0^1βtan^(β1)β‘π₯/(1 + π₯^2 ) ππ₯ Step 1 : Let F(π₯)=β«1βtan^(β1)β‘π₯/(1+γ π₯γ^2 ) ππ₯ Put tan^(β1)β‘π₯=π‘ Differentiating w.r.t.π₯ π/ππ₯ (tan^(β1)β‘π₯ )=ππ‘/ππ₯ 1/(1 + π₯^2 )=ππ‘/ππ₯ Therefore, β«1βtan^(β1)β‘π₯/(1+γ π₯γ^2 ) ππ₯=β«1βγπ‘/(1+π₯^2 ) Γ (1+π₯^2 )ππ‘γ =β«1βγπ‘ ππ‘γ =π‘^2/2 Putting π‘=γπ‘ππγ^(β1)β‘π₯ =(tan^(β1)β‘π₯ )^2/2 Hence πΉ(π₯)=(tan^(β1)β‘π₯ )^2/2 Step 2 : β«1βγπ‘ππγ^(β1)β‘π₯/(1 + π₯^2 )=πΉ(1)βF(0) =1/2 (tan^(β1)β‘1 )^2 β1/2 (tan^(β1)β‘0 )^2 =1/2 (π/4)^2β1/2 (0)^2 =1/2 π^2/16 = π ^π/ππ Example 27 (Method 2) Evaluate β«_0^1βtan^(β1)β‘π₯/(1 + π₯^2 ) ππ₯ Put π‘=tan^(β1)β‘π₯ Differentiating w.r.t.π₯ ππ‘/ππ₯=π/ππ₯ (tan^(β1)β‘π₯ ) ππ‘/ππ₯=1/(1 + π₯^2 ) (1+π₯^2 )ππ‘=ππ₯ Hence when value of x varies from 0 to 1, value of t varies from 0 to π/4 Therefore, β«_0^1βtan^(β1)β‘π₯/(1 + π₯^2 )=β«_0^(π/4)βπ‘/(1 + π₯^2 ) ππ₯ (1+π₯^2 )ππ‘ =β«_0^(π/4)βγ π‘ ππ‘γ =[π‘^2/2]_0^(π/4) =1/2 [(π/4)^2β(0)^2 ] =1/2 Γ π^2/16 = π ^π/ππ