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Evaluate \[\int\frac{1-\sin x}{\cos^2x}\,dx.\]

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Question

Evaluate \[\int\frac{1-\sin x}{\cos^2x}\,dx.\]

Options

  • \[\tan x+\sec x+C\]

  • \[-\tan x-\sec x+C\]

  • \[\tan x-\sec x+C\]

  • \[-\cot x-\operatorname{cosec}x+C\]

MCQ
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Solution

Write the integrand as \[\sec^2x-\tan x\sec x.\] Integrating gives \[\tan x-\sec x+C,\] because \[\int\sec^2x\,dx=\tan x\] and \[\int\tan x\sec x\,dx=\sec x.\]

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