Misc 19 - Chapter 7 Class 12 Integrals
Last updated at April 16, 2024 by Teachoo
Miscellaneous
Misc 1
Important
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Chapter 7 Class 12 Integrals
Serial order wise
Transcript
Misc 19 Integrate the function β((1 β βπ₯)/(1 + βπ₯)) β«1βγβ((1 β βπ₯)/(1 + βπ₯)) ππ₯γ Let x = γπππγ^π ππ½ dx = β4 cos 2π sin 2π dπ Substituting, = β«1ββ((1 β β((γπππ γ^2 2π) ))/(1 + β((γπππ γ^2 2π) )))Γβ4 cosβ‘2ΞΈ sinβ‘2ΞΈ ππ = β«1ββ((1 β cosβ‘2π)/(1 + πππ 2π))Γ(β4) cosβ‘2ΞΈ sinβ‘2ΞΈ ππ = β4β«1ββ((1 β (1 β 2γπ ππγ^(2 ) π))/(1 + (2γπππ γ^(2 ) π β 1) )) cosβ‘2ΞΈ (2 sinβ‘ΞΈ cosβ‘γπ)γ ππ = β8β«1ββ((2γπ ππγ^(2 ) π)/(2γπππ γ^(2 ) π)) cosβ‘2ΞΈ cosβ‘π sinβ‘π ππ = β8β«1βsinβ‘π/cosβ‘ΞΈ cosβ‘2ΞΈ cosβ‘π sinβ‘π ππ = β8β«1βγγπ ππγ^2 πγ cosβ‘2ΞΈ ππ = β8β«1β((1 β cosβ‘2ΞΈ)/2) cosβ‘2ΞΈ ππ = β4 β«1β(πππ 2ΞΈβcos^2β‘2ΞΈ ) ππ = 4 β«1β(γπππ γ^2 2ΞΈβcosβ‘2ΞΈ ) ππ = 4 β«1βγγπππ γ^2 2ΞΈγ ππβ4β«1βcosβ‘2ΞΈ ππ = 4 β«1β(cosβ‘4π + 1)/2 ππ β 4β«1βγπππ 2πγ ππ = 2 β«1βγ(cosβ‘4π + 1)γ ππ β 4β«1βγπππ 2πγ ππ = 2 [(sinβ‘4 π)/4+π] β4 [(sinβ‘2 π)/2]+C = sinβ‘4π/2+2π β2 π ππ 2π+ C Now x = γπππ γ^2 2π βπ₯ " = " cosβ‘2π γπππ γ^(β1) βπ₯=2π 1/2 γπππ γ^(β1) βπ₯=π And, sin 4π = 2 sin 2π cos 2π = 2β(1βπ₯)Γβπ₯ = 2 βπ₯ β(1βπ₯) Putting the values. = sinβ‘4π/2+2ΞΈβ2 sinβ‘2ΞΈ+ C = (2βπ₯ β(1 β π₯))/2+2 (γπππ γ^(β1) βπ₯)/2β2β(1βπ₯)+C = βπ₯ β(1βπ₯)+γπππ γ^(β1) βπ₯β2β(1βπ₯)+ C = β(π₯βπ₯^2 )+γπππ γ^(β1) βπ₯β2β(1βπ₯)+C = β2β(πβπ)+γπππγ^(βπ) βπ+β(πβπ^π )+π
