Definite Integration - By Substitution

Chapter 7 Class 12 Integrals
Concept wise

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### Transcript

Misc 27 Evaluate the definite integral โซ_0^(๐/2)โใ(cos^2โก๐ฅ ๐๐ฅ)/(cos^2โก๐ฅ + 4 sin^2โก๐ฅ ) ใ Let I = โซ1_0^(๐/2)โ(ใ๐๐๐ ใ^2 ๐ฅ)/(ใ๐๐๐ ใ^2 ๐ฅ + 4ใ๐ ๐๐ใ^2 ๐ฅ) ๐๐ฅ = โซ1_0^(๐/2)โ(ใ๐๐๐ ใ^2 ๐ฅ)/(ใ๐๐๐ ใ^2 ๐ฅ + 4(ใ1 โ ๐๐๐ ใ^2 ๐ฅ) ๐๐ฅ) = โซ1_0^(๐/2)โ(ใ๐๐๐ ใ^2 ๐ฅ)/(4 โ 3 ใ๐๐๐ ใ^2 ๐ฅ) ๐๐ฅ = (โ1)/3 โซ1_0^(๐/4)โใ (โ3 ใ๐๐๐ ใ^2 ๐ฅ )/(4 โ 3 ใ๐๐๐ ใ^2 ๐ฅ) ใ ๐๐ฅ = (โ1)/3 โซ1_0^(๐/2)โ(ใโ 3 ๐๐๐ ใ^2 ๐ฅ + 4 โ 4)/(4 โ 3 ใ๐๐๐ ใ^2 ๐ฅ) ๐๐ฅ = (โ1)/3 โซ1_0^(๐/2)โ(ใ4 โ 3 ๐๐๐ ใ^2 ๐ฅ โ 4)/(4 โ 3 ใ๐๐๐ ใ^2 ๐ฅ) ๐๐ฅ = (โ1)/3 โซ1_0^(๐/2)โใ1โ4/(4 โ 3 ใ๐๐๐ ใ^2 ๐ฅ)ใ ๐๐ฅ Dividing numerator and denominator by ใ๐๐๐ ใ^2 ๐ฅ = (โ1)/3 (๐/2)+4/3 โซ1_0^(๐/2)โ(๐๐ฅ/(ใ๐๐๐ ใ^2 ๐ฅ))/((4 โ 3 ใ๐๐๐ ใ^2 ๐ฅ )/(ใ๐๐๐ ใ^2 ๐ฅ)) ๐๐ฅ = (โ1)/3 (๐/2)+4/3 โซ1_0^(๐/2)โ(ใ๐ ๐๐ใ^2 ๐ฅ)/(4 ใ๐ ๐๐ใ^2 ๐ฅ โ 3) ๐๐ฅ = (โ๐)/6+4/3 โซ1_0^(๐/2)โ(ใ๐ ๐๐ใ^2 ๐ฅ)/(4 (1 + ใ๐ก๐๐ใ^2 ๐ฅ) โ 3) ๐๐ฅ = (โ๐)/6+4/3 โซ1_0^(๐/2)โ(ใ๐ ๐๐ใ^2 ๐ฅ)/(4 + 4 ใ๐ก๐๐ใ^2 ๐ฅ โ 3) ๐๐ฅ = (โ๐)/6+4/3 โซ1_0^(๐/2)โ(ใ๐ ๐๐ใ^2 ๐ฅ)/( 4 ใ๐ก๐๐ใ^2 ๐ฅ + 1) ๐๐ฅ Let tan x = t Differentiating w.r.t x ใ๐ ๐๐ใ^2 x dx = dt Thus, When x = 0, t = 0, & when x = ๐/2, ๐ก= โ Substituting values and limit I = (โ๐)/6+4/3 โซ1_0^(๐/2)โ(ใ๐ ๐๐ใ^2 ๐ฅ)/( 4 ใ๐ก๐๐ใ^2 ๐ฅ + 1) ๐๐ฅ โด I =(โ๐)/6+4/3 โซ1_0^โโ๐๐ก/(ใ4๐กใ^2+1) = (โ๐)/6+4/3 โ1/4 โซ1_0^โโ๐๐ก/(ใ ๐กใ^2+1/4) =(โ๐)/6+4/3โ1/4 ร 1/((1/2) ) [ใ๐ก๐๐ใ^(โ1) ๐ก/(1/2)]_0^โ = (โ๐)/6+2/3โ [ใ๐ก๐๐ใ^(โ1) 2๐ก]_0^โ =(โ๐)/6+2/3โใ[๐ก๐๐ใ^(โ1) โโใ๐ก๐๐ใ^(โ1) 0] = โ๐/6+2/3โ[๐/2โ0] =โ๐/6+๐/3 =๐/๐ We know that โซ1โ1/(๐ฅ^2 + ๐^2 )=ใ1/๐ tan^(โ1)ใโกใ๐ฅ/๐ใ+๐ถ = (โ๐)/6+2/3โ[๐/2โ0] =(โ๐)/6+๐/3 =๐/3โ๐/6 =๐/๐