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The value of 0πsin(n+1/2)xsin(x/2)dx(nN) is

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detailed solution

Correct option is A

We have,2sin⁡x212+cos⁡x+cos⁡2x+…+cos⁡nx=sin⁡x2+2sin⁡x2cos⁡x+2sin⁡x2cos⁡2x+…+2sin⁡x2,cos nx=sin⁡x2+sin⁡3x2−sin⁡x2+sin⁡5x2−sin⁡3x2+…+sin⁡n+12x−sin⁡n−12x=sin⁡n+12x12+cos⁡x+cos⁡2x+…+cos⁡nx=sin⁡n+12x2sin⁡(x/2)⇒ ∫0π sin⁡n+12xsin⁡(x/2)dx=2∫0π 12dx+∫0π cos⁡xdx+…+∫0π cos⁡nxdx=2π2+sin⁡x0π+…+sin⁡nxn0π=π


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