मराठी

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

Advertisements
Advertisements

प्रश्न

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

पर्याय

  • \[\tan x+\sec x+C\]

  • \[-\tan x-\sec x+C\]

  • \[\tan x-\sec x+C\]

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

MCQ
Advertisements

उत्तर

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 \[\int\tan x\sec x\,dx=\sec x.\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×