If $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sqrt{1-\sin 2x} dx = \alpha + \beta\sqrt{2} + \gamma\sqrt{3}$, where $\alpha, \beta$ and $\gamma$ are rational numbers, then $3\alpha + 4\beta - \gamma$ is equal to ______.

JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Definite Integrals

\int_{\frac{π}{6}}^{\frac{π}{3}} 1 - sin 2 x d x = α + β 2 + γ 3

, where

α, β

and

γ

are rational numbers, then

3 α + 4 β - γ

is equal to ______.

Answer: 6

1 - sin 2 x = sin^{2} x + cos^{2} x - 2 sin x cos x = (sin x - cos x)^{2}

, so

1 - sin 2 x = ∣ sin x - cos x ∣

.

On

[\frac{π}{6}, \frac{π}{3}]

sin x - cos x

changes sign at

x = \frac{π}{4}

: negative before it, positive after.

\int_{π /6}^{π /3} 1 - sin 2 x d x = \int_{π /6}^{π /4} (cos x - sin x) d x + \int_{π /4}^{π /3} (sin x - cos x) d x

First piece:

[sin x + cos x]_{π /6}^{π /4} = (\frac{2}{2} + \frac{2}{2}) - (\frac{1}{2} + \frac{3}{2}) = 2 - \frac{1}{2} - \frac{3}{2}

.

Second piece:

[- cos x - sin x]_{π /4}^{π /3} = (- \frac{1}{2} - \frac{3}{2}) - (- 2) = 2 - \frac{1}{2} - \frac{3}{2}

.

Sum

= 22 - 1 - 3

, so

α = - 1, β = 2, γ = - 1

3 α + 4 β - γ = - 3 + 8 + 1 = 6

JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Integer answer