Miscellaneous

Chapter 7 Class 12 Integrals
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Misc 31 Evaluate the definite integral β«_0^(π/2)βγsinβ‘2π₯ tan^(β1)β‘(sinβ‘π₯ ) γ ππ₯ β«_0^(π/2)βγsinβ‘2π₯ tan^(β1)β‘(sinβ‘π₯ ) γ ππ₯ = β«_0^(π/2)βγ2 sinβ‘π₯ cosβ‘π₯ tan^(β1)β‘(sinβ‘π₯ ) γ ππ₯ Let sinβ‘π₯=π‘ Differentiating both sides π€.π.π‘.π₯ cosβ‘π₯=ππ‘/ππ₯ ππ₯=ππ‘/cosβ‘π₯ Substituting x and dx β«1_0^(π/2)βγ2 sinβ‘γπ₯ cosβ‘γπ₯ γπ‘ππγ^(β1) (sinβ‘γπ₯) γ γ γ γ ππ₯ = β«1_0^1βγ2π‘ cosβ‘γπ₯ γπ‘ππγ^(β1) (π‘) γ γ ππ‘/πππ π₯ = β«1_0^1βγ2π‘ γπ‘ππγ^(β1) π‘ γ ππ‘ = 2β«1_0^1βγπ‘ γπ‘ππγ^(β1) (π‘) γ ππ‘ =2(γπ‘ππγ^(β1) π‘β«1βπ‘ ππ‘ β β«1βγ((π (γπ‘ππγ^(β1) ))/ππ‘ β«1βγπ‘ ππ‘ γ) γ ππ‘) Now we know, β«1βγπ (π₯) π(π₯) γ ππ₯=π(π₯) β«1βπ(π₯) ππ₯ββ«1βγ((π π (π₯))/ππ₯ β«1βγπ(π₯) ππ₯γ) γ ππ₯ Putting f(x) = γπ‘ππγ^(β1) (π‘) & g(x) = t = 2(γπ‘ππγ^(β1) π‘ (γπ‘/2γ^2 )ββ«1β1/(1 + π‘^2 )Γπ‘^2/2 ππ‘) = 2(γπ‘/2γ^2 γπ‘ππγ^(β1) π‘β1/2 β«1βπ‘^2/2 ππ‘) = π‘^2 γπ‘ππγ^(β1) π‘ββ«1βπ‘^2/(1 + π‘^2 ) ππ‘ Solving π°_π I_1 = β«1βπ‘^2/(1 + π‘^2 ) ππ‘ Adding and Subtracting 1 in numerator. I_1 = β«1β((π‘^2 + 1 β 1)/(π‘^2 + 1))ππ‘ I_1= β«1β((π‘^2 + 1)/(π‘^2 + 1)β1/(π‘^2 + 1))ππ‘ I_1= β«1β(1β1/(π‘^2 + 1)) ππ‘ I_1= β«1βγππ‘ββ«1βππ‘/(π‘^2 + 1)γ I_1= t β γπ‘ππγ^(β1) (t) Thus, our equation becomes β΄ β«1βγγπ‘ππγ^(β1) (π‘)Γπ‘ ππ‘γ= π‘^2 γπ‘ππγ^(β1) π‘ βI_1 = π‘^2 γπ‘ππγ^(β1) π‘β(π‘βγπ‘ππγ^(β1) π‘) = π‘^2 γπ‘ππγ^(β1) π‘βπ‘+γπ‘ππγ^(β1) π‘ Now, 2β«1_0^1βγγπ‘ππγ^(β1) (π‘) π‘ ππ‘γ =[π‘^2 γπ‘ππγ^(β1) π‘βπ‘+γπ‘ππγ^(β1) π‘]_0^1 =(1^2Γγπ‘ππγ^(β1) 1β1+γπ‘ππγ^(β1) 1)β(0β0+γπ‘ππγ^(β1) (0)) =(π/4β 1+π/4)β0 = π/πβπ