Find ∫e x   √(1 + sin ⁑2x )/(1 + cos ⁑2x ) dx

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Question 9 Find ∫1▒〖𝑒^π‘₯ √(1 + sin⁑2π‘₯ )/(1 + cos⁑2π‘₯ )γ€— dx ∫1▒〖𝑒^π‘₯ √(1 + 𝑠𝑖𝑛 2π‘₯ )/(1 + π‘π‘œπ‘  2π‘₯)γ€— dx Putting sin 2x = 2 sin x cos x cos 2x = 2 cos2 x – 1 = ∫1▒〖𝑒^π‘₯ √(1 + 2 sin⁑π‘₯ cos⁑π‘₯ )/(1 + (2 cos^2⁑π‘₯ βˆ’ 1))γ€— dx = ∫1▒〖𝑒^π‘₯ √(1 + 2 sin⁑π‘₯ cos⁑π‘₯ )/(2 cos^2⁑π‘₯ )γ€— dx Putting 1 = sin2 x + cos2 x = ∫1▒〖𝑒^π‘₯ √(sin^2⁑π‘₯ + cos^2⁑π‘₯ + 2 sin⁑π‘₯ cos⁑π‘₯ )/(2 cos^2⁑π‘₯ )γ€— dx = ∫1β–’βˆš((𝑠𝑖𝑛 π‘₯ + π‘π‘œπ‘  π‘₯)^2 )/(2 π‘π‘œπ‘ 2π‘₯)dx = 1/2 ∫1▒〖𝑒^π‘₯ (sin⁑π‘₯/(π‘π‘œπ‘ 2 π‘₯)+cos⁑π‘₯/(π‘π‘œπ‘ 2 π‘₯))𝑑π‘₯γ€— = 1/2 ∫1▒〖𝑒π‘₯ ("sec sec tan x " )𝑑π‘₯γ€— = 1/2 𝑒^π‘₯ sec x + c [βˆ΅π‘’^π‘₯ (𝑓(π‘₯)+𝑓^β€² (π‘₯))𝑑π‘₯=𝑒^π‘₯ 𝑓(π‘₯)+𝑐] Using (a + b)2 = a2 + b2 + 2ab where a = sin x , b = cos x = ∫1▒〖𝑒^π‘₯ √((𝑠𝑖𝑛 π‘₯ + π‘π‘œπ‘  π‘₯)^2 )/(2 π‘π‘œπ‘ 2π‘₯)γ€—dx = 1/2 ∫1▒〖𝑒^π‘₯ ((𝑠𝑖𝑛 π‘₯ + π‘π‘œπ‘  π‘₯))/π‘π‘œπ‘ 2π‘₯γ€—dx = 1/2 ∫1▒〖𝑒^π‘₯ (sin⁑π‘₯/(π‘π‘œπ‘ 2 π‘₯)+cos⁑π‘₯/(π‘π‘œπ‘ 2 π‘₯))𝑑π‘₯γ€— = 1/2 ∫1▒〖𝑒^π‘₯ (sin⁑π‘₯/(π‘π‘œπ‘  π‘₯) Γ—1/cos⁑π‘₯ +1/cos⁑π‘₯ )𝑑π‘₯γ€— = 1/2 ∫1▒〖𝑒^π‘₯ (tan⁑π‘₯ sec⁑π‘₯+sec⁑π‘₯ )𝑑π‘₯γ€— = 1/2 ∫1▒〖𝑒^π‘₯ (sec⁑π‘₯+tan⁑π‘₯ sec⁑π‘₯ )𝑑π‘₯γ€— Using ∫1▒〖𝑒^π‘₯ (𝑓(π‘₯)+𝑓^β€² (π‘₯))𝑑π‘₯γ€—=𝑒^π‘₯ 𝑓(π‘₯)+𝑐 Where f(x) = sec x = 𝟏/𝟐 𝒆^𝒙 sec x + c

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