Q31 / 104 · NTA Abhyas (JEE Main) 2020 · Mathematics · Indefinite Integration

The integral ∫ x  cos  - 1 1 - x 2 1 + x 2 dx;  x > 0 is equal to

  • A
    - x + 1 + x 2  cot - 1 x + c
  • B
    x - 1 + x 2  cot - 1 x + c
  • C
    x - 1 + x 2  tan - 1 x + c
  • Correct answer
    D
    - x + 1 + x 2  tan - 1 x + c

Solution

I = ∫ x cos -1 1 - x 2 1 + x 2 dx;  (x > 0)

Let,  x  = tan θ ⇒  dx  = sec 2 θ d θ

cos -1 1 - x 2 1 + x 2 = cos -1 cos 2 θ

                            = 2 θ

                            = 2 tan -1 x

I = ∫ tan θ 2 θ · sec 2 θ  d θ

    = 2 θ ∫ tan θ sec 2 θ d θ - ∫ ∫ tan θ sec 2 θ d θ d d θ θ d θ

    = 2 θ tan 2 θ 2 - ∫ tan 2 θ 2 1 d θ

    = 2 2 θ tan 2 θ - ∫ sec 2 θ - 1 d θ

    =θ tan 2 θ - tan θ + θ + c

    = x 2 tan -1 x - x + tan -1 x + c

    = - x + 1 + x 2 tan -1 x + c

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

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