Misc 26 - Chapter 7 Class 12 Integrals (Term 2)
Last updated at Dec. 23, 2019 by Teachoo
Transcript
Misc 26 Evaluate the definite integral β«_0^(π/4)βγ(sinβ‘π₯ cosβ‘π₯)/(cos^4β‘π₯ + sin^4β‘π₯ ) ππ₯γ β«_0^(π/4)βγ(sinβ‘π₯ cosβ‘π₯)/(cos^4β‘π₯ + sin^4β‘π₯ ) ππ₯γ Adding and subtracting 2 γπ ππγ^2 π₯ γπππ γ^2 π₯ from denominator = β«_0^(π/4)βγ(sinβ‘π₯ cosβ‘π₯)/(cos^4β‘π₯ + sin^4β‘γπ₯ + 2γπ ππγ^2 π₯ γπππ γ^2 π₯ β 2γπ ππγ^2 π₯ γπππ γ^2 π₯γ ) ππ₯γ = β«_0^(π/4)βγ(sinβ‘π₯ cosβ‘π₯)/(γγ(cosγ^2β‘π₯ + sin^2β‘π₯)γ^2 β 2γπ ππγ^2 π₯ γπππ γ^2 π₯) ππ₯γ = β«_0^(π/4)βγ(sinβ‘π₯ cosβ‘π₯)/(1 β (4 γπ ππγ^2 π₯ γπππ γ^2 π₯)/2) ππ₯γ = β«_0^(π/4)βγ(sinβ‘π₯ cosβ‘π₯)/(1 β 1/2 (2 sinβ‘π₯ cosβ‘π₯ )^2 ) ππ₯γ = β«_0^(π/4)βγ(sinβ‘π₯ cosβ‘π₯)/(1 β (γπ ππγ^2 2π₯)/2) ππ₯γ = β«_0^(π/4)βγ(2 sinβ‘π₯ cosβ‘π₯)/(2 β γπ ππγ^(2 ) 2π₯) ππ₯γ = β«_0^(π/4)βγsinβ‘2π₯/(1 + (1 β γπ ππγ^(2 ) 2π₯)) ππ₯γ = β«_0^(π/4)βγsinβ‘2π₯/(1 + γπππ γ^2 2π₯) γ ππ₯ (γπππ γ^2 π₯+γπ ππγ^2 π₯=1" " ) ("2 sin " π₯ cosβ‘π₯=sinβ‘2π₯ " " ) ("2 sin " π₯ cosβ‘π₯=sinβ‘2π₯ " " ) Let t = cos 2π₯ ππ‘/ππ₯=β2 sinβ‘2π₯ (βππ‘)/2=sinβ‘2π₯ ππ₯ So, our equation becomes β«_0^(π/4)βγsinβ‘2π₯/(1 + γπππ γ^2 2π₯) γ ππ₯ = (β1)/2 β«1_1^0βππ‘/(1 + π‘^2 ) = (β1)/2 [γπ‘ππγ^(β1) (π‘)]_1^0 = (β1)/2 [γπ‘ππγ^(β1) (0)βγπ‘ππγ^(β1) (1)] = (β1)/2 (0βπ/4) = π /π
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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.