Q29 / 104 · NTA Abhyas (JEE Main) 2020 · Mathematics · Indefinite Integration

The value of the integral ∫e3sin-1x11-x2+e3cos-1xdx is equal to
(where, c is an arbitrary constant)

  • A
    e3sin-1x3+xe3π2+c
  • B
    esin-1x+eπ/2+c
  • Correct answer
    C
    e3sin-1x3+xe3π2+c
  • D
    eπ2+exπ2+c

Solution

∫e3sin-1x1-x2dx+∫e3sin-1x+cos-1xdx
=∫e3tdt+∫e3π2dx Put sin-1⁡x=t
=e3t3+e3π2⋅x+c

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

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