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

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

Answered by MathematicalUser2357 last updated on 18/Dec/24

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

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