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a
logsint+12+sin2t+sint+c
b
logsint+12+sin2t−sint+c
c
logsint+sin2t+sint+c
d
logsint−12+sin2t+sint+c

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detailed solution

Correct option is A

∫cosect−1dt=∫1−sintsintdt=∫1−sint1+sintsint(1+sint)dt Use =∫1y2−a2=log⁡y+y2−a2+c=∫cos⁡tsin2⁡t+sin⁡tdt Put sin⁡t=y=∫dyy2+y=∫dyy+122−122+c=log⁡y+12+y2+y+c=log⁡sin⁡t+12+sin2⁡t+sin⁡t+c


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