# Misc 26

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 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 β2γπ 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)^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 + (1β γπ ππγ^(2 ) 2π₯)) ππ₯γ = β«_0^(π/4)βγsinβ‘2π₯/(1 + γπππ γ^(2 ) 2π₯) γ dβ‘π₯ Let t = cos 2π₯ ππ‘/ππ₯=β2 sinβ‘2π₯ (βππ‘)/2=sinβ‘2π₯ ππ₯ Substituting value and changing limits = (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

Misc 1

Misc 2

Misc 3

Misc 4

Misc 5

Misc 6

Misc 7

Misc 8 Important

Misc 9

Misc 10

Misc 11

Misc 12

Misc 13

Misc 14

Misc 15

Misc 16

Misc 17

Misc 18 Important

Misc 19 Important

Misc 20 Important

Misc 21

Misc 22

Misc 23

Misc 24 Important

Misc 25

Misc 26 You are here

Misc 27

Misc 28

Misc 29

Misc 30 Important

Misc 31

Misc 32 Important

Misc 33

Misc 34

Misc 35

Misc 36

Misc 37

Misc 38

Misc 39

Misc 40

Misc 41 Important

Misc 42

Misc 43

Misc 44 Important

Integration Formula Sheet - Chapter 7 Class 12 Formulas Important

About the Author

CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .