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Q.

∫cos⁡5x+cos⁡4x1−2cos⁡3xdx 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=∫cos⁡5x+cos⁡4x1−2cos⁡3xdx=∫2cos⁡9x2cos⁡x21−22cos2⁡3x2−1dx=∫2cos⁡9x2cos⁡x23−4cos2⁡3x2dx=∫2cos⁡9x2cos⁡x2cos⁡3x23cos⁡3x2−4cos3⁡3x2dx=∫2cos⁡9x2cos⁡3x2cos⁡x2−cos⁡9x2dx=-∫2cos⁡3x2cos⁡x2dx=−∫(cos⁡2x+cos⁡x)dx=−sin⁡2x2−sin⁡x+c
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