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B
12log(2+sinx2−sinx)+C
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C
14log(1+sinx1−sinx)+C
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D
None of the above
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Solution
The correct option is A14log(2−sinx+cosx2+sinx−cosx)+C Let I=∫sinx+cosx3+sin2xdx =∫sinx+cosx4+sin2x−1dx ⇒I=∫sinx+cosx4−(sinx−cosx)2dx Put sinx−cosx=t ⇒(sinx+cosx)dx=dt Therefore, I=∫dt4−t2=14log2−t2+t+C =14log(2−sinx+cosx2+sinx−cosx)+C