Simplify the integrand
We use
1 − sin 2 x = sin 2 x + cos 2 x − 2 sin x cos x = ( sin x − cos x ) 2 . 1-\sin 2x = \sin^2 x + \cos^2 x - 2\sin x\cos x = (\sin x-\cos x)^2. 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 ∣ . \sqrt{1-\sin 2x}=|\sin x-\cos x|. 1 − sin 2 x = ∣ sin x − cos x ∣. On the interval
x ∈ [ π 6 , π 3 ] , x\in\left[\frac{\pi}{6},\frac{\pi}{3}\right], x ∈ [ 6 π , 3 π ] , sin x − cos x \sin x-\cos x sin x − cos x
At x = π 6 x=\frac{\pi}{6} x = 6 π sin π 6 − cos π 6 = 1 2 − 3 2 < 0. \sin\frac{\pi}{6}-\cos\frac{\pi}{6}=\frac12-\frac{\sqrt3}{2}<0. sin 6 π − cos 6 π = 2 1 − 2 3 < 0.
At x = π 3 x=\frac{\pi}{3} x = 3 π sin π 3 − cos π 3 = 3 2 − 1 2 > 0. \sin\frac{\pi}{3}-\cos\frac{\pi}{3}=\frac{\sqrt3}{2}-\frac12>0. sin 3 π − cos 3 π = 2 3 − 2 1 > 0.
The sign changes at x = π 4 x=\frac{\pi}{4} x = 4 π sin x = cos x \sin x=\cos x sin x = cos x
Hence,
1 − sin 2 x = { cos x − sin x , π 6 ≤ x ≤ π 4 , sin x − cos x , π 4 ≤ x ≤ π 3 . \sqrt{1-\sin 2x}=
\begin{cases}
\cos x-\sin x, & \frac{\pi}{6}\le x\le \frac{\pi}{4},\\[4pt]
\sin x-\cos x, & \frac{\pi}{4}\le x\le \frac{\pi}{3}.
\end{cases} 1 − sin 2 x = { cos x − sin x , sin x − cos x , 6 π ≤ x ≤ 4 π , 4 π ≤ x ≤ 3 π .
Split the integral
Therefore,
I = ∫ π / 6 π / 3 1 − sin 2 x d x = ∫ π / 6 π / 4 ( cos x − sin x ) d x + ∫ π / 4 π / 3 ( sin x − cos x ) d x . I=\int_{\pi/6}^{\pi/3}\sqrt{1-\sin 2x}\,dx
=\int_{\pi/6}^{\pi/4}(\cos x-\sin x)\,dx+\int_{\pi/4}^{\pi/3}(\sin x-\cos x)\,dx. I = ∫ π /6 π /3 1 − sin 2 x d x = ∫ π /6 π /4 ( cos x − sin x ) d x + ∫ π /4 π /3 ( sin x − cos x ) d x .
Evaluate the first integral
∫ ( cos x − sin x ) d x = sin x + cos x . \int (\cos x-\sin x)dx=\sin x+\cos x. ∫ ( cos x − sin x ) d x = sin x + cos x . So,
I 1 = [ sin x + cos x ] π / 6 π / 4 . I_1=\left[\sin x+\cos x\right]_{\pi/6}^{\pi/4}. I 1 = [ sin x + cos x ] π /6 π /4 . Now,
sin π 4 + cos π 4 = 2 2 + 2 2 = 2 , \sin\frac{\pi}{4}+\cos\frac{\pi}{4}=\frac{\sqrt2}{2}+\frac{\sqrt2}{2}=\sqrt2, sin 4 π + cos 4 π = 2 2 + 2 2 = 2 , sin π 6 + cos π 6 = 1 2 + 3 2 . \sin\frac{\pi}{6}+\cos\frac{\pi}{6}=\frac12+\frac{\sqrt3}{2}. sin 6 π + cos 6 π = 2 1 + 2 3 . Hence,
I 1 = 2 − 1 2 − 3 2 . I_1=\sqrt2-\frac12-\frac{\sqrt3}{2}. I 1 = 2 − 2 1 − 2 3 .
Evaluate the second integral
∫ ( sin x − cos x ) d x = − cos x − sin x . \int (\sin x-\cos x)dx=-\cos x-\sin x. ∫ ( sin x − cos x ) d x = − cos x − sin x . So,
I 2 = [ − cos x − sin x ] π / 4 π / 3 . I_2=\left[-\cos x-\sin x\right]_{\pi/4}^{\pi/3}. I 2 = [ − cos x − sin x ] π /4 π /3 . Now,
− cos π 3 − sin π 3 = − 1 2 − 3 2 , -\cos\frac{\pi}{3}-\sin\frac{\pi}{3}=-\frac12-\frac{\sqrt3}{2}, − cos 3 π − sin 3 π = − 2 1 − 2 3 , − cos π 4 − sin π 4 = − 2 . -\cos\frac{\pi}{4}-\sin\frac{\pi}{4}=-\sqrt2. − cos 4 π − sin 4 π = − 2 . Therefore,
I 2 = ( − 1 2 − 3 2 ) − ( − 2 ) = 2 − 1 2 − 3 2 . I_2=\left(-\frac12-\frac{\sqrt3}{2}\right)-(-\sqrt2)
=\sqrt2-\frac12-\frac{\sqrt3}{2}. I 2 = ( − 2 1 − 2 3 ) − ( − 2 ) = 2 − 2 1 − 2 3 .
Add both parts
I = I 1 + I 2 = 2 ( 2 − 1 2 − 3 2 ) = 2 2 − 1 − 3 . I=I_1+I_2=2\left(\sqrt2-\frac12-\frac{\sqrt3}{2}\right)
=2\sqrt2-1-\sqrt3. I = I 1 + I 2 = 2 ( 2 − 2 1 − 2 3 ) = 2 2 − 1 − 3 . Thus,
α = − 1 , β = 2 , γ = − 1. \alpha=-1,\quad \beta=2,\quad \gamma=-1. α = − 1 , β = 2 , γ = − 1.
Compute the required value
3 α + 4 β − γ = 3 ( − 1 ) + 4 ( 2 ) − ( − 1 ) = − 3 + 8 + 1 = 6. 3\alpha+4\beta-\gamma=3(-1)+4(2)-(-1)=-3+8+1=6. 3 α + 4 β − γ = 3 ( − 1 ) + 4 ( 2 ) − ( − 1 ) = − 3 + 8 + 1 = 6. So the required integer is
6 . \boxed{6}. 6 .
Comparison with stored answer
Stored correct answer = 6 6 6
Our derived answer also equals 6 6 6