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Question

The value of π20sin3xsinx+cosxdx is:
  1. π12
  2. π14 
  3.  π28
  4. π24


Solution

The correct option is B π14 
I=π/20sin3xsinx+cosxdx(i)Also, I=π/20sin3(π2x)sin(π2x)+cos(π2x)dx=π/20cos3xsinx+cosxdx(ii)
On adding (i) and (ii) we get
2I=π/20sin3x+cos3xsinx+cosxdxI=12π20sin3x+cos3xsinx+cosxdxI=12π/20(1sinxcosx)dx=12[xsin2x2]π/20
=π414=π14
 

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