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Mathematics Question on Calculus

If∫π6π31−sin⁡2x dx=α+β2+γ3,\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sqrt{1 - \sin 2x} \, dx = \alpha + \beta \sqrt{2} + \gamma \sqrt{3},where α\alpha, β\beta, and γ\gamma are rational numbers, then 3α+4β−γ3\alpha + 4\beta - \gamma is equal to ______.

Answer

Step 1. Evaluate the integral:
∫π8π31−sin⁡2x dx\int_{\frac{\pi}{8}}^{\frac{\pi}{3}} \sqrt{1 - \sin 2x} \, dx

Step 2. Rewrite 1−sin⁡2x\sqrt{1 - \sin 2x} using trigonometric identities:
∫π8π3∣sin⁡x−cos⁡x∣ dx\int_{\frac{\pi}{8}}^{\frac{\pi}{3}} |\sin x - \cos x| \, dx

Step 3. Split the integral based on the intervals where sin⁡x−cos⁡x\sin x - \cos x changes sign:

=∫π8π4(cos⁡x−sin⁡x) dx+∫π4π3(sin⁡x−cos⁡x) dx= \int_{\frac{\pi}{8}}^{\frac{\pi}{4}} (\cos x - \sin x) \, dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} (\sin x - \cos x) \, dx

Step 4. Solve each integral:
=−1+22−3= -1 + 2\sqrt{2} - \sqrt{3}
Thus, we have:
α+β2+γ3=−1+22−3\alpha + \beta \sqrt{2} + \gamma \sqrt{3} = -1 + 2\sqrt{2} - \sqrt{3}

where α=−1\alpha = -1, β=2\beta = 2, and γ=−1\gamma = -1.

Step 5. Calculate 3α+4β−γ3\alpha + 4\beta - \gamma:
3α+4β−γ=3(−1)+4(2)−(−1)=−3+8+1=63\alpha + 4\beta - \gamma = 3(-1) + 4(2) - (-1) = -3 + 8 + 1 = 6

The Correct Answer is: 3α+4β−γ=63\alpha + 4\beta - \gamma = 6.