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The value of is
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a
14logsinx−1sinx+1−12log2sinx−12sinx+1+C
b
18logcosx−1cosx+1−122log2cosx−12cosx+1+C
c
18logsinx−1sinx+1−142log2sinx−12sinx+1+C
d
none of these
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Correct option is C
The given integrand can be written as 14cosxcos2x=cosx41−sin2x1−2sin2x, hence∫sinxsin4xdx=14∫dt1−t21−2t2 (t=sinx)=14∫1t2−1−1t2−1/2dt=1412logt−1t+1−12logt−1/2t+1/2+C=18logsinx−1sinx+1−142log2sinx−12sinx+1+C.Similar Questions
is equal to
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