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intcos.(x)/(16)cos.(x)/(8)cos.(x)/(4)cos...

`intcos.(x)/(16)cos.(x)/(8)cos.(x)/(4)cos.(x)/(2)sin.(x)/(16)dx=`

A

(a) `1/(16)cosx+c`

B

(b) `(-1)/(16)cosx+c`

C

(c) `1/(16)sinx+c`

D

(d) `(-1)/4cos""x/2+c`

Text Solution

Verified by Experts

The correct Answer is:
D
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