Question ID: #575
If $24\int_{0}^{\frac{\pi}{4}}\left(\sin\left|4x-\frac{\pi}{12}\right|+[2 \sin x]\right)dx=2\pi+\alpha$, where $[.]$ denotes the greatest integer function, then $\alpha$ is equal to
Solution:
Let $I = \int_{0}^{\frac{\pi}{4}}\left(\sin\left|4x-\frac{\pi}{12}\right|+[2 \sin x]\right)dx$.
First, analyze the modulus function $|4x – \frac{\pi}{12}|$.
The critical point is $4x = \frac{\pi}{12} \Rightarrow x = \frac{\pi}{48}$.
So we split the interval at $\frac{\pi}{48}$.
Next, analyze the greatest integer function $[2 \sin x]$.
For $x \in [0, \frac{\pi}{4}]$, $0 \le \sin x \le \frac{1}{\sqrt{2}} \approx 0.707$.
Thus, $0 \le 2\sin x \le \sqrt{2} \approx 1.414$.
$[2 \sin x] = 0$ when $0 \le 2\sin x < 1 \Rightarrow 0 \le \sin x < \frac{1}{2} \Rightarrow 0 \le x < \frac{\pi}{6}$. $[2 \sin x] = 1$ when $1 \le 2\sin x \le \sqrt{2} \Rightarrow \frac{\pi}{6} \le x \le \frac{\pi}{4}$.
Now, split the integral $I$ into parts:
$$ I = \int_{0}^{\frac{\pi}{48}} \sin\left(\frac{\pi}{12}-4x\right) dx + \int_{\frac{\pi}{48}}^{\frac{\pi}{4}} \sin\left(4x-\frac{\pi}{12}\right) dx + \int_{0}^{\frac{\pi}{6}} 0 \, dx + \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} 1 \, dx $$
Calculate the first integral:
$$ \int_{0}^{\frac{\pi}{48}} \sin\left(\frac{\pi}{12}-4x\right) dx = \left[ \frac{\cos(\frac{\pi}{12}-4x)}{4} \right]_{0}^{\frac{\pi}{48}} $$
$$ = \frac{1}{4} \left( \cos(0) – \cos\left(\frac{\pi}{12}\right) \right) = \frac{1}{4} \left( 1 – \cos\frac{\pi}{12} \right) $$
Calculate the second integral:
$$ \int_{\frac{\pi}{48}}^{\frac{\pi}{4}} \sin\left(4x-\frac{\pi}{12}\right) dx = \left[ -\frac{\cos(4x-\frac{\pi}{12})}{4} \right]_{\frac{\pi}{48}}^{\frac{\pi}{4}} $$
$$ = -\frac{1}{4} \left( \cos\left(\pi – \frac{\pi}{12}\right) – \cos(0) \right) = -\frac{1}{4} \left( -\cos\frac{\pi}{12} – 1 \right) = \frac{1}{4} \left( 1 + \cos\frac{\pi}{12} \right) $$
Calculate the fourth integral:
$$ \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} 1 \, dx = \frac{\pi}{4} – \frac{\pi}{6} = \frac{3\pi – 2\pi}{12} = \frac{\pi}{12} $$
Total Sum $I$:
$$ I = \frac{1}{4} \left( 1 – \cos\frac{\pi}{12} \right) + \frac{1}{4} \left( 1 + \cos\frac{\pi}{12} \right) + \frac{\pi}{12} $$
$$ I = \frac{1}{4}(2) + \frac{\pi}{12} = \frac{1}{2} + \frac{\pi}{12} $$
The given equation is $24 I = 2\pi + \alpha$.
$$ 24 \left( \frac{1}{2} + \frac{\pi}{12} \right) = 12 + 2\pi $$
$$ 2\pi + 12 = 2\pi + \alpha $$
$$ \alpha = 12 $$
Ans. 12
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