Miscellaneous

Chapter 7 Class 12 Integrals
Serial order wise

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### Transcript

Misc 36 Prove that β«_0^(π/4)βγ2 tanγ^3β‘π₯ ππ₯=1βlogβ‘2 Solving L.H.S 2β«_0^(π/4)βγ tanγ^3β‘π₯ ππ₯ = 2β«_0^(π/4)βγ tan π₯ tanγ^2β‘π₯ ππ₯ = 2β«_0^(π/4)βγ tan π₯ (secγ^2β‘γπ₯β1)γ ππ₯ = 2β«_0^(π/4)βγ tan π₯ secγ^2β‘π₯ ππ₯β 2β«_0^(π/4)βtanβ‘γπ₯ ππ₯γ π°_π = 2β«_π^(π/π)βγ π­ππ§ π πππγ^πβ‘π ππ Let t = tan x ππ‘/ππ₯ = γπ ππγ^2 x dt = γπ ππγ^2x dx Substituting, 2β«1_0^1βγπ‘ ππ‘γ = 2 [π‘^2/2]_0^1 = 2 (1/2β0) =1 π°_π= 2β«_π^(π/π)βπππβ‘γπ ππγ 2β«_π^(π/π)βπππβ‘γπ ππγ = 2 ["log" |saecβ‘π₯ |]_0^(π/4) = 2 ("log" β2β0) = 2 (log 2^(1/2)) = 2 (1/2 "log " 2) = log 2 Hence, = πΌ_1βπΌ_2 = 1 β log 2 = R.H.S Hence, proved.