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JEE Main 2019 · 9 Jan · Shift I

JEE Main 2019 Mathematics question on Definite Integrals and Areas

The value of ∫−ππ∣cos⁡x∣3 dx\int_{-\pi}^{\pi} |\cos x|^3 \, dx is
  1. A.
    23\frac{2}{3}
  2. B.
    −43-\frac{4}{3}
  3. C.
    00
  4. D.
    43\frac{4}{3}
Show answer and solution

Correct option: D

Solution

Since |cos x|³ is even, the integral equals 2∫₀^π |cos x|³ dx. On [0,π], cos x is non-negative on [0,π/2] and non-positive on [π/2,π], so |cos x|³ = cos³x on [0,π/2] and -cos³x... wait, |cos x|³ = |cos x|³ is always positive. By symmetry about π/2, 2∫₀^π |cos x|³ dx = 4∫₀^(π/2) cos³x dx.

Now ∫₀^(π/2) cos³x dx = ∫₀^(π/2) (1-sin²x)cos x dx = [sin x - sin³x/3]₀^(π/2) = 1 - 1/3 = 2/3.

So the total = 4 × (2/3) = 4/3.

[JEE Main 2019 · 9 Jan · Shift I]

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