Question-219549

Question Number 219549 by Spillover last updated on 28/Apr/25

Answered by Nicholas666 last updated on 29/Apr/25

∫((1+sin x)/(1−sinx))dx=2tany−x+c ∫(((1+sin x)^2 )/(1−sin^2 x))dx=2 tan y −x +c ∫((1+2 sin x+sin^2 x)/(cos^2 x))dx=2 tan y −x+c ∫(sec^2 x+2 sec x tan^2 x )dx=2 tan y −x + c ∫(sec^2 x+2 sec x tan x+sec^2 x − 1)dx=2 tan y−x+c ∫(2sec^2 x+2 sec x tan x−1)dx=2 tan y−x+c 2 tan x −2 sec x −x + c = 2 tan y−x + c 2 tanx −2 sec x = 2 tan y tan y = tan x + sec x Final Answer; tan y = tan x + sec x

$$\int\frac{\mathrm{1}+{sin}\:{x}}{\mathrm{1}−{sinx}}{dx}=\mathrm{2}{tany}−{x}+{c} \\ $$$$\int\frac{\left(\mathrm{1}+{sin}\:{x}\right)^{\mathrm{2}} }{\mathrm{1}−{sin}^{\mathrm{2}} {x}}{dx}=\mathrm{2}\:{tan}\:{y}\:−{x}\:+{c} \\ $$$$\int\frac{\mathrm{1}+\mathrm{2}\:{sin}\:{x}+{sin}^{\mathrm{2}} {x}}{{cos}^{\mathrm{2}} {x}}{dx}=\mathrm{2}\:{tan}\:{y}\:−{x}+{c} \\ $$$$\int\left({sec}^{\mathrm{2}} {x}+\mathrm{2}\:{sec}\:{x}\:{tan}^{\mathrm{2}} \:{x}\:\right){dx}=\mathrm{2}\:{tan}\:{y}\:−{x}\:+\:{c} \\ $$$$\int\left({sec}^{\mathrm{2}} {x}+\mathrm{2}\:{sec}\:{x}\:{tan}\:{x}+{sec}^{\mathrm{2}} {x}\:−\:\mathrm{1}\right){dx}=\mathrm{2}\:{tan}\:{y}−{x}+{c} \\ $$$$\int\left(\mathrm{2}{sec}^{\mathrm{2}} {x}+\mathrm{2}\:{sec}\:{x}\:{tan}\:{x}−\mathrm{1}\right){dx}=\mathrm{2}\:{tan}\:{y}−{x}+{c}\:\:\: \\ $$$$\mathrm{2}\:{tan}\:{x}\:−\mathrm{2}\:{sec}\:{x}\:−{x}\:+\:{c}\:=\:\mathrm{2}\:{tan}\:{y}−{x}\:+\:{c} \\ $$$$\mathrm{2}\:{tanx}\:−\mathrm{2}\:{sec}\:{x}\:=\:\mathrm{2}\:{tan}\:{y} \\ $$$${tan}\:{y}\:=\:{tan}\:{x}\:+\:{sec}\:{x} \\ $$$$ \\ $$$$\mathrm{Final}\:\mathrm{Answer}; \\ $$$$\:\:\:\:\:\:\:\:\:{tan}\:{y}\:=\:{tan}\:{x}\:+\:{sec}\:{x} \\ $$$$ \\ $$

Commented by Spillover last updated on 01/May/25