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a
(8sinx−1)6433−8sinx+12+3364sin−18sinx−133+C
b
8sinx−16433+8sinx−12+3364sin−18sinx−133+C
c
8sinx−16433−8sinx−12+3364sin−18sinx−133+C
d
8sinx−1648sinx−12−33+3364sin−18sinx−133+C
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Correct option is C
Let I=∫2cos2x+sinxcosxdx=∫21−2sin2x+sinxcosxdx=−4sin2x+sinx+2cosxdx Put sinx=t=∫−4t2+t+2dt=∫(−4)t2−14t−24dt=∫(−4)t−182−182−12dt=∫(−4)t−182−3364dt=2∫3382−t−182 dt=2t−1823382−t−182+33822sin−1t−18338+C=28t−1163364−(8t−1)642+33128sin−18t−133+CSimilar Questions
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