Miscellaneous

Chapter 7 Class 12 Integrals
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Question 1 Integrate the function (sin^(β1)β‘βπ₯ β cos^(β1)β‘βπ₯)/(sin^(β1)β‘βπ₯ + cos^(β1)β‘βπ₯ ) , π₯β[0, 1] Let πΌ = β«1β(sin^(β1)β‘βπ₯ β cos^(β1)β‘βπ₯)/(sin^(β1)β‘βπ₯ + cos^(β1)β‘βπ₯ ) ππ₯ We can write as (sin^(β1)β‘βπ₯ β cos^(β1)β‘βπ₯)/(sin^(β1)β‘βπ₯ + cos^(β1)β‘βπ₯ ) = (sin^(β1)β‘βπ₯ β (π/2 " β" γ π ππγ^(β1)β‘βπ₯ ))/(π/2) We know that γπ ππγ^(β1)β‘π₯+γπππ γ^(β1)β‘π₯=π/2 or γπππ γ^(β1)β‘π₯=π/2 βγ π ππγ^(β1)β‘π₯ = (2/π)(sin^(β1)β‘βπ₯ βπ/2 " +" γ π ππγ^(β1)β‘βπ₯ ) = 2/π (2 sin^(β1)β‘βπ₯ βπ/2) = 2/π Γ2 sin^(β1)β‘βπ₯β 2/πΓπ/2 = 4/π sin^(β1)β‘βπ₯β1 Integrating π€.π.π‘.π₯ β«1β(sin^(β1)β‘βπ₯ β cos^(β1)β‘βπ₯)/(sin^(β1)β‘βπ₯ + cos^(β1)β‘βπ₯ ) ππ₯=β«1β(4/π sin^(β1)β‘βπ₯β1) ππ₯ = β«1βγ4/π sin^(β1)β‘βπ₯ γ ππ₯ββ«1βππ₯ = 4/π β«1βsin^(β1)β‘βπ₯ ππ₯βπ₯+πΆ1 Let πΌ1=β«1βsin^(β1)β‘βπ₯ ππ₯ Hence, I = 4/π πΌ1βπ₯+πΆ1 Solving π_π πΌ1 = β«1βsin^(β1)β‘βπ₯ ππ₯ Put βπ₯=π‘ π₯=π‘^2 Differentiating π€.π.π‘.π₯ ππ₯/ππ₯ = (ππ‘^2)/ππ₯ 1 = 2π‘ ππ‘/ππ₯ ππ₯ = 2π‘ ππ‘ Therefore β«1βsin^(β1)β‘βπ₯ ππ₯=β«1βsin^(β1)β‘π‘ .2π‘ ππ‘ =2β«1βsin^(β1)β‘π‘ .π‘ ππ‘ =2β«1βγπ‘ sin^(β1)β‘γπ‘ γ γ ππ‘ =2[sin^(β1)β‘γπ‘ γ β«1βπ‘ ππ‘ββ«1β((π/ππ‘ sin^(β1)β‘π‘ ) β«1βγπ‘ ππ‘γ) ππ‘ Now we know that β«1βγπ(π₯) πβ‘(π₯) γ ππ₯=π(π₯) β«1βπ(π₯) ππ₯ββ«1β(π^β² (π₯) β«1βπ(π₯) ππ₯) ππ₯ Putting f(x) = t and g(x) = sinβ1 t =2[sin^(β1)β‘γπ‘ γ π‘^2/2 ββ«1β1/β(1 βγ π‘γ^2 ) Γπ‘^2/2 ππ‘+πΆ] =2Γπ‘^2/2 γ sin^(β1)γβ‘π‘β2Γβ«1βγ1/2 Γπ‘^2/β(1 βγ π‘γ^2 )γ ππ‘+πΆ = π‘^2 sin^(β1)β‘π‘ββ«1βπ‘^2/β(1 βγ π‘γ^2 ) ππ‘+πΆ = π‘^2 sin^(β1)β‘π‘+β«1β(βπ‘^2)/β(1 βγ π‘γ^2 ) ππ‘+πΆ Solving β«1βγβ πγ^π/β(π βγ πγ^π ) ππ We can write (β π‘^2)/β(1 βγ π‘γ^2 ) =(γβ π‘γ^2 + 1 β 1)/β(1 βγ π‘γ^2 ) =(γ1 β π‘γ^2 β 1)/β(1 βγ π‘γ^2 ) =γ1 β π‘γ^2/β(1 βγ π‘γ^2 ) β" " 1/β(1 βγ π‘γ^2 ) =β(1 βγ π‘γ^2 ) β" " 1/β(1 βγ π‘γ^2 ) Integrating π€.π.π‘.π₯ β«1β(β π‘^2)/(1 βγ π‘γ^2 ) dt = β«1βγ(β(1 βγ π‘γ^2 ) β" " 1/β(1 βγ π‘γ^2 )) γ ππ‘ = β«1ββ(1^2 βγ π‘γ^2 ) ππ‘ββ«1β1/β(1^2 βγ π‘γ^2 ) ππ‘ = π‘/2 β(1^2 βγ π‘γ^2 )+1^2/2 sin^(β1)β‘γπ‘/1γβsin^(β1)β‘γπ‘/1γ We know that β«1ββ(π^2βπ₯^2 )=π₯/2 β(π^2βπ₯^2 )+π^2/2 sin^(β1)β‘γπ₯/πγ+πΆ β«1β1/β(π^2 β π₯^2 )=sin^(β1)β‘γπ₯/πγ+πΆ = π‘/2 β(1 βγ π‘γ^2 )+1/2 sin^(β1)β‘π‘βsin^(β1)β‘π‘ = π‘/2 β(1 βγ π‘γ^2 ) β 1/2 sin^(β1)β‘π‘ Hence we can write πΌ1 = π‘^2 sin^(β1)β‘π‘+β«1β(β π‘^2)/β(1 βγ π‘γ^2 ) ππ‘ πΌ1 = π‘^2 sinβ‘π‘+π‘/2 β(1 βγ π‘γ^2 )β1/2 sin^(β1)β‘π‘ Putting π‘ = βπ₯ πΌ1 = (βπ₯)^2 sinβ‘βπ₯+βπ₯/2 β(1 βγ (βπ₯)γ^2 )β1/2 sin^(β1)β‘βπ₯ πΌ1 = π₯ sinβ‘βπ₯+βπ₯/2 β(1βπ₯)β1/2 sin^(β1)β‘βπ₯ Hence πΌ = 4/π γ πΌγ_(1 )βπ₯+C_1 πΌ = 4/π (π₯ γπ ππγ^(β1) βπ₯+βπ₯/2 β(1βπ₯)β1/2 γπ ππγ^(β1) βπ₯)βπ₯+C_1 πΌ = 4/π (π₯ γπ ππγ^(β1) βπ₯+β(π₯ β π₯^2 )/2 β1/2 γπ ππγ^(β1) βπ₯)βπ₯+C_1 πΌ = 4/π π₯ γπ ππγ^(β1) βπ₯+2/π β(π₯ β π₯^2 )β2/π γπ ππγ^(β1) βπ₯βπ₯+C_1 πΌ = 4/π π₯ γπ ππγ^(β1) βπ₯β2/π γπ ππγ^(β1) βπ₯+2/π β(π₯ β π₯^2 )βπ₯+C_1 πΌ = γπ ππγ^(β1) βπ₯ [4π₯/πβ2/π]+(2 β(π₯ β π₯^2 ))/πβπ₯+ C_1 πΌ = γπ ππγ^(β1) βπ₯ [(4π₯ β 2)/π]+(2 β(π₯ β π₯^2 ))/πβπ₯+ C_1 π° = γπππγ^(βπ) βπ [(π(ππ βπ))/π]+(π β(π β π^π ))/πβπ+ πͺ_π