# Misc 38 - Chapter 7 Class 12 Integrals (Term 2)

Last updated at Dec. 23, 2019 by Teachoo

Miscellaneous

Misc 1
Important

Misc 2 Important

Misc 3 Important

Misc 4

Misc 5 Important

Misc 6

Misc 7 Important

Misc 8 Important

Misc 9

Misc 10 Important

Misc 11

Misc 12

Misc 13

Misc 14 Important

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 Important

Misc 26 Important

Misc 27 Important

Misc 28 Important

Misc 29

Misc 30 Important

Misc 31 Important

Misc 32 Important

Misc 33 Important

Misc 34

Misc 35

Misc 36

Misc 37

Misc 38 Important You are here

Misc 39

Misc 40 Important Deleted for CBSE Board 2022 Exams

Misc 41 (MCQ) Important

Misc 42 (MCQ)

Misc 43 (MCQ)

Misc 44 (MCQ) Important

Integration Formula Sheet - Chapter 7 Class 12 Formulas Important

Chapter 7 Class 12 Integrals (Term 2)

Serial order wise

Misc 38 Prove that β«_0^(π/4)βγ2 tanγ^3β‘π₯ ππ₯=1βlogβ‘2 Solving L.H.S 2β«_0^(π/4)βγ tanγ^3β‘π₯ ππ₯ = 2β«_0^(π/4)βγ tan π₯ tanγ^2β‘π₯ ππ₯ = 2β«_0^(π/4)βγ tan π₯ (secγ^2β‘γπ₯β1)γ ππ₯ = 2β«_0^(π/4)βγ tan π₯ secγ^2β‘π₯ ππ₯β 2β«_0^(π/4)βtanβ‘γπ₯ ππ₯γ π°_π = 2β«_π^(π /π)βγ πππ§ π πππγ^πβ‘π π π Let t = tan x ππ‘/ππ₯ = γπ ππγ^2 x dt = γπ ππγ^2x dx Substituting, 2β«1_0^1βγπ‘ ππ‘γ = 2 [π‘^2/2]_0^1 = 2 (1/2β0) =1 π°_π= 2β«_π^(π /π)βπππβ‘γπ π πγ 2β«_π^(π /π)βπππβ‘γπ π πγ = 2 ["log" |saecβ‘π₯ |]_0^(π/4) = 2 ("log" β2β0) = 2 (log 2^(1/2)) = 2 (1/2 "log " 2) = log 2 Hence, = πΌ_1βπΌ_2 = 1 β log 2 = R.H.S Hence, proved.