NTA Abhyas JEE Main2020MathematicsIndefinite IntegrationPractice

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

Options

A- x + 1 + x 2 cot - 1 x + c
Bx - 1 + x 2 cot - 1 x + c
Cx - 1 + x 2 tan - 1 x + c
D- x + 1 + x 2 tan - 1 x + c

Correct answer

D. - x + 1 + x 2 tan - 1 x + c

Step-by-step 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

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