If $$\int_\limits{\frac{\pi}{6}}^{\frac{\pi}{3}} \sqrt{1-\sin 2 x} d x=\alpha+\beta \sqrt{2}+\gamma \sqrt{3}$$, where $\alpha, \beta$ and $\gamma$ are rational numbers, then $3 \alpha+4 \beta-\gamma$ is equal to _________.
Answer (integer)
6
Solution
<p>$$\begin{aligned}
& =\int_\limits{\frac{\pi}{6}}^{\frac{\pi}{3}} \sqrt{1-\sin 2 x} d x \\
& =\int_\limits{\frac{\pi}{6}}^{\frac{\pi}{3}}|\sin x-\cos x| d x \\
& =\int_\limits{\frac{\pi}{6}}^{\frac{\pi}{4}}(\cos x-\sin x) d x+\int_\limits{\frac{\pi}{4}}^{\frac{\pi}{3}}(\sin x-\cos x) d x \\
& =-1+2 \sqrt{2}-\sqrt{3} \\
& =\alpha+\beta \sqrt{2}+\gamma \sqrt{3} \\
& \alpha=-1, \beta=2, \gamma=-1 \\
& 3 \alpha+4 \beta-\gamma=6
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Definite Integration · Topic: Fundamental Theorem of Calculus
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