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The value of integral $I=∫\frac{1-sinx}{cos^2x}dx$ is : |
tan x - sec x + c, where c is a constant sec x - tan x + c, where c is a constant sec x tan x + c, where c is a constant tan x + sec x + c, where c is a constant |
tan x - sec x + c, where c is a constant |
The correct answer is Option (1) → $\tan x - \sec x + c$, where c is a constant $I=∫\frac{1-\sin x}{\cos^2x}dx$ $=∫\frac{1}{\cos^2x}-\frac{\sin x}{\cos x}×\frac{1}{\cos x}dx$ $=∫\sec^2x-\sec x\tan xdx$ $=\tan x-\sec x+c$ |
