The value of ∫ - π 2 π 2 x 2 cos ⁡ x 1 + e x   d x is equal to

The value of ∫-π2π2x2cos⁡x1+ex dx is equal to
  1. π24-2
  2. π24+2
  3. π2-eπ2
  4. π2+eπ2

Solution

Let I= ∫ −  x 2 x 2 x 2 cosx 1+ e x dx
Replacing x by -x
I= ∫ − π 2 π 2 ( x 2 cosx 1+ e −x )dx
⇒2I= ∫ − π 2 π 2 ( 1 1+ e x + 1 1+ e −x ) x 2 cosx dx
⇒2I= ∫ − π 2 π 2 x 2 cosx dx , (even function)
I = ∫0π2x2cos⁡x dx=x2sin⁡x0π2-2∫0π2x.sin⁡x dx
(Using integration by parts, taking x 2 =u,cos x dx = dv )
=π24-2-x.cos⁡x0π2+ ∫0π21.cos⁡xdx=π24-20+1=π24-2

Asked in: JEE Advanced 2016 (Paper 2)

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