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Question Number 214568 by Ikbal last updated on 12/Dec/24

∫((b+ax)/(1+sin x)) dx

$$\int\frac{{b}+{ax}}{\mathrm{1}+\mathrm{sin}\:{x}}\:{dx} \\ $$

Answered by MathematicalUser2357 last updated on 18/Dec/24

∫((ax+b)/(sin x+1))dx =∫(((ax+b)sec x)/(tan x+sec x))dx =−((ax+b)/(tan x+sec x))+∫(a/(tan x+sec x))dx =−((ax+b)/(tan x+sec x))+a∫(1/(tan x+sec x))dx =−((ax+b)/(tan x+sec x))+a∫((sec x−tan x)/(sec^2 x−tan^2 x))dx =−((ax+b)/(tan x+sec x))+∫(sec x−tan x)dx =−((ax+b)/(tan x+sec x))+∫sec xdx−∫tan xdx =−((ax+b)/(tan x+sec x))+ln∣tan x+sec x∣dx−ln∣cos x∣+C

$$\int\frac{{ax}+{b}}{\mathrm{sin}\:{x}+\mathrm{1}}{dx} \\ $$$$=\int\frac{\left({ax}+{b}\right)\mathrm{sec}\:{x}}{\mathrm{tan}\:{x}+\mathrm{sec}\:{x}}{dx} \\ $$$$=−\frac{{ax}+{b}}{\mathrm{tan}\:{x}+\mathrm{sec}\:{x}}+\int\frac{{a}}{\mathrm{tan}\:{x}+\mathrm{sec}\:{x}}{dx} \\ $$$$=−\frac{{ax}+{b}}{\mathrm{tan}\:{x}+\mathrm{sec}\:{x}}+{a}\int\frac{\mathrm{1}}{\mathrm{tan}\:{x}+\mathrm{sec}\:{x}}{dx} \\ $$$$=−\frac{{ax}+{b}}{\mathrm{tan}\:{x}+\mathrm{sec}\:{x}}+{a}\int\frac{\mathrm{sec}\:{x}−\mathrm{tan}\:{x}}{\mathrm{sec}^{\mathrm{2}} {x}−\mathrm{tan}^{\mathrm{2}} {x}}{dx} \\ $$$$=−\frac{{ax}+{b}}{\mathrm{tan}\:{x}+\mathrm{sec}\:{x}}+\int\left(\mathrm{sec}\:{x}−\mathrm{tan}\:{x}\right){dx} \\ $$$$=−\frac{{ax}+{b}}{\mathrm{tan}\:{x}+\mathrm{sec}\:{x}}+\int\mathrm{sec}\:{xdx}−\int\mathrm{tan}\:{xdx} \\ $$$$=−\frac{{ax}+{b}}{\mathrm{tan}\:{x}+\mathrm{sec}\:{x}}+\mathrm{ln}\mid\mathrm{tan}\:{x}+\mathrm{sec}\:{x}\mid{dx}−\mathrm{ln}\mid\mathrm{cos}\:{x}\mid+{C} \\ $$

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