Q39 / 90 · JEE Main 2018 (08 Apr) · Mathematics

The integral ∫sin2xcos2xsin5x+cos3xsin2x+sin3xcos2x+cos5x2dx, is equal to

(where C is the constant of integration).

  • A
    -11+cot3x+C
  • B
    131+tan3x+C
  • Correct answer
    C
    -131+tan3x+C
  • D
    11+cot3x+C

Solution

∫sin2xcos2xsin5x+cos3xsin2x+sin3xcos2x+cos5x2dx

=∫sin2xcos2xsin3xsin2x+cos2x+cos3x cos2x+sin2x2dx 

=∫sin2xcos2xsin3x+cos3x2dx

=∫sin2xcos2xcos3x2tan3x+12dx

=∫tan2xsec2xtan3x+12dx
Put, tan3x+1=t
⇒dt=3tan2xsec2xdx
=∫dt3t2=-13t+C, where C is the constant of integration.
=  -13tan3x+1+C

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

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