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JEE Main 2019 · 12 Jan · Shift I

JEE Main 2019 Mathematics question on Indefinite Integration

The integral ∫cos⁡(log⁡ex) dx\int \cos(\log_e x)\, dx is equal to (where CC is a constant of integration)
  1. A.
    x2[cos⁡(log⁡ex)+sin⁡(log⁡ex)]+C\frac{x}{2}[\cos(\log_e x) + \sin(\log_e x)] + C
  2. B.
    x[cos⁡(log⁡ex)+sin⁡(log⁡ex)]+Cx[\cos(\log_e x) + \sin(\log_e x)] + C
  3. C.
    x[cos⁡(log⁡ex)−sin⁡(log⁡ex)]+Cx[\cos(\log_e x) - \sin(\log_e x)] + C
  4. D.
    x2[sin⁡(log⁡ex)−cos⁡(log⁡ex)]+C\frac{x}{2}[\sin(\log_e x) - \cos(\log_e x)] + C
Show answer and solution

Correct option: A

Solution

Use integration by parts twice. Let I = ∫cos(ln x)dx.

First IBP: u = cos(ln x), dv = dx → I = x·cos(ln x) + ∫sin(ln x)dx.

Second IBP on ∫sin(ln x)dx: u = sin(ln x), dv = dx → ∫sin(ln x)dx = x·sin(ln x) - ∫cos(ln x)dx = x·sin(ln x) - I.

Substituting back: I = x·cos(ln x) + x·sin(ln x) - I, so 2I = x[cos(ln x) + sin(ln x)], giving I = (x/2)[cos(ln x) + sin(ln x)] + C.

[JEE Main 2019 · 12 Jan · Shift I]

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