Q2 / 104 · NTA Abhyas (JEE Main) 2020 · Mathematics · Indefinite Integration

The integral I=∫2sin⁡x3+sin⁡2xdx simplifies to (where, C is the constant of integration)

  • A
    ln⁡2+sin⁡x-cos⁡x2-sin⁡x+cos⁡x-tan-1⁡sin⁡x+cos⁡x+C
  • B
    ln⁡(sin⁡x)+sin⁡2x+C
  • C
    sin⁡2x-ln⁡(cos⁡x)+C
  • Correct answer
    D
    14ln⁡2+sin⁡x-cos⁡x2-sin⁡x+cos⁡x-12tan-1⁡sin⁡x+cos⁡x2+C

Solution

The given integral can be rewritten as,

I=∫sin⁡x+cos⁡x3+sin⁡2xdx+∫sin⁡x-cos⁡x3+sin⁡2xdx
Let sin⁡x-cos⁡x=t and sin⁡x+cos⁡x=u in 1st and 2nd integral respectively.
∴I=∫dt4-t2-∫du2+u2
=14ln⁡2+t2-t-12tan-1⁡u2+C
Putting values of t and u, we get,

I=14ln⁡2+sinx-cosx2-sinx+cosx-12tan-1⁡sinx+cosx2+C

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

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