Q.
∫cos5x+cos4x1−2cos3xdx is equal to
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a
sin2x2+cosx+cundefined
b
sin2x2−cosx+c
c
−sin2x2−sinx+c
d
sin2x2+cosx+c
answer is C.
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Detailed Solution
I=∫cos5x+cos4x1−2cos3xdx=∫2cos9x2cosx21−22cos23x2−1dx=∫2cos9x2cosx23−4cos23x2dx=∫2cos9x2cosx2cos3x23cos3x2−4cos33x2dx=∫2cos9x2cos3x2cosx2−cos9x2dx=-∫2cos3x2cosx2dx=−∫(cos2x+cosx)dx=−sin2x2−sinx+c
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