Q43 / 90 · JEE Main 2019 (08 Apr Shift 1) · Mathematics

∫sin5x2sinx2dx, is equal to

  • Correct answer
    A
    x+2sinx+sin2x+c
  • B
    2x+sinx+sin2x+c
  • C
    x+2sinx+2sin2x+c
  • D
     2x+sinx+2sin2x+c

Solution

We have, I=∫sin5x2sinx2dx  

=∫sin2x+x2sinx2dx

Using sinA+B=sinAcosB+cosAsinB

=∫sin2xcosx2+cos2xsinx2sinx2dx

=∫2 sinx cosxcosx2sinx2+cos 2xdx

=∫22sinx2cosx2cosx cosx2sinx2+cos 2xdx

=∫4cos2x2cosx+cos2xdx

Using  cos2x=2cos2x-1⇒2cos2x=1+cos2x

=∫21+cosxcosx+cos2xdx

=∫2cosx+2cos2x+cos2xdx

=∫2cosx+1+cos2x+cos2xdx

=∫2cosx+2cos2x+1dx

=2sinx+sin2x+x+c, where c is the constant of integration.

Hence, the correct answer is option: 'A'.

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