Miscellaneous

Chapter 7 Class 12 Integrals
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Misc 22 Integrate the function tan^(β1)β‘β((1 β π₯)/(1 + π₯)) Let x = cos 2π ππ₯/ππ=β2 sinβ‘γ2π γ dx = β2 sin 2π dπ Substituting, β«1βγtan^(β1)β‘β((1 β π₯)/(1 + π₯)) ππ₯γ = β«1βγπ‘ππγ^(β1) β((1 β cosβ‘2π)/(1 + cosβ‘2π ))Γ(β2 sinβ‘γ2 π)γ π π = β2β«1βγπ‘ππγ^(β1) β((1 β (1 β 2γπ ππγ^2 π))/(1 + (2γπππ γ^2 π β 1) ))Γ("sin 2π dπ " ) = β2β«1βγπ‘ππγ^(β1) β((sin^2β‘π/cos^2β‘π ) )Γsinβ‘γ2π ππγ = β2β«1βγπ‘ππγ^(β1) (sinβ‘π/cosβ‘π )Γsinβ‘γ2π ππγ = β2β«1βγπ‘ππγ^(β1) (π‘ππβ‘π )Γsinβ‘γ2π ππγ = β 2 β«1βπ sinβ‘γ2π ππγ =β2(πβ«1βγsinβ‘2π ππβ(β«1βπ(π)/ππ β«1βsinβ‘2π ππ) ππγ) =β2(π((βcosβ‘2π)/2)ββ«1β1((βcosβ‘2π)/2) ππ) =β2(βπ(cosβ‘2π/2)β€Ά+β«1βcosβ‘2π/2 ππ) =β2(β(π cosβ‘2π)/2β€Ά+sinβ‘2π/4) 1/2 γπππ γ^(β1) (π₯)=π π = 1/2 γπππ γ^(β1) π₯ π₯^2=γπππ γ^2 2π π₯^2=1βγπ ππγ^2 2π γπ ππγ^2 2π="1 β " π₯^2 sin 2π = β(1βπ₯^2 ) Now, x = cos 2π Putting the values = β2 (β1/2 (1/2 γπππ γ^(β1) π₯)π₯+β(1 β π₯^2 )/4) = β2 (β(1 β π₯^2 )/4β(π₯ γπππ γ^(β1) π₯)/4)+ C = π/π (πγ πππγ^(βπ) πββ(πβπ^π ) )+ C