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cosxcos3x1cos3xdx is equal to 

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a
23sin−1⁡cos3/2⁡x+C
b
32sin−1⁡cos3/2⁡x+C
c
23cos−1⁡cos3/2⁡x+C
d
None of these

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detailed solution

Correct option is C

∫cos⁡x−cos3⁡x1−cos3⁡xdx=∫cos⁡x1−cos3⁡xsin⁡xdx=∫t1−t3dt=−∫t1−t3/22dt, where t=cos⁡x=−23∫32t1−t3/22dt=23cos−1⁡t3/2+C=23cos−1⁡(cos⁡x)3/2+C


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